## Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 9 Sequences and Series

**Short Answer** **Type Questions **

**(Score 3)**

Question 1.

Which term of the A.P. 4,9,14, …is 109?

a. P+2,4P-6,3P-2 are three consecutive terms of an A.P. Find the value of P.

Answer:

a.Given a = 4, d = 5,a =109

⇒ 4 + (n-1)5 = 109,n-1 =21,n = 22

22^{nd} term of the given A.P is 109

b. P + 2 + 3p-2 = 8p-12 4p = 8p -12,4p = 12, p = 3

Question 2.

a The third term of a GP is 3. The product of first 5 term is

a. 1024

b. 243

c. 3125

d. 32

b. The 5th term of a GP is 2, then find the product of its first 9 terms.

Answer:

Question 3.

Find the 20th term of GP 5/2,5/4,5/8, ……………..

Answer:

Question 4.

a A sequence containing a finite number of terms is called

b. If the 8^{th} term of an A.P is zero. Prove that its 28^{th} term is double the 18^{th} term.

Answer:

a. finite sequence

b. Given a_{g} = 0, ⇒ a + 7d = 0, a = -7d

a_{28} = a + 27d = 20d

a_{28} = a+ 17d= 10d

a_{28} = 2 ×a_{18}

Question 5.

If p and q are positive real numbers such that p ^{2} + q^{2} = 1, then find the maximum value of(p + q).

Answer:

Question 6.

Find the sum of the series 2^{2} + 4^{2} + 6^{2} +…. + (2n)^{2}. Also find 2^{2} + 4^{2} + 6^{2} +……….. + 48^{2}.

Answer:

Question 7.

If the sum of three numbers are in GP. is 38 and their product is 1728. Find the numbers.

Answer:

Question 8.

Insert 3 geometric means between the numbers 1 and 256.

Answer:

Question 9.

Find the middle terms in the A.P 20,16,12, …,-176

Answer:

a = 20, d = -4, a — -176

⇒ a + (n-1) d = -176

20 + (n-1) -4 = -176

n-1 =49, n = 50

Hence the middle terms are 25^{th} and 26^{th }a + 24d = 20 + -96 = -76

a + 25d = 20 + -100 = -80

Question 10.

a. If a, b, care in AP and also

are in AP, then

a. a = b≠c

b. a=b=c

c. a = b = c

d. a A b = c

b. 1^{3} + 2^{3} + 3^{3} +…………. 10^{3} + = …………….

Answer:

**Short Answer** **Type Questions **

**(Score 4)**

Question 1.

a A student read common difference of an AP as -2 instead of 2 and got the sum of first 5 terms as -5. Actual sum of first 5 terms is

a. 25

b. 35

c. -35

d. -25

b. Which term of the GP-2,4,-8,16,. is 256

Answer:

⇒ 5×7 = 35

b. Given a =-2, r=-2, a =256

⇒ ar^{n-1} = 256, -2 (-2)-^{1} = 256

(-2)^{n} = (-2)^{8} n= 8

256 is the 8^{th} term of the given GP.

Question 2.

a. n^{th} term of some sequence are given below. Which term can be the n^{th} term of an A.P?

i. a_{n}=n(n+1)

ii. a_{n} = 2 + 5n

iii. a_{n} = 2^{2} + 2

iv. a_{n} = n^{2} + n +1

b. If the sum of 12^{th} and 22^{rd} terms of an A.P is 100. Find the sum of first 33 terms.

Answer:

Question 3.

a. If the sum 24 + 20 +16 +……. to n terms =72. Find n.

b. Find the sum of first 35 terms of an A.P, if second term = 2 and seventh term=22

Answer:

i. Given S_{n} = 72 ⇒ [48 + (n -1) – 4] =72

n[48-4n + 4] = 144

4n^{2}-52n+ 144 = 0

n^{2}– 13n+36 = 0

(n – 9) (n – 4) = 0, n = 9 or n = 4

ii. a + d = 2

a+6d = 22 5d = 20, d = 4, a = -2

Question 4.

If the sum of p terms of an A.Pis same as the sum of its q terms. Show that the sum of its (p+q) terms is zero.

Answer:

Question 5.

If the 4^{th} and 9^{th} terms of a GP are 54 and 13122 respectively. Find the GP. h. In a GP, whose 8^{th} term is 192 and common ratio is 2. Find 1^{st} term and 12^{th} term.

Answer:

Question 6.

a. The sum of three numbers in A.P is -3 and their product is 8. Find the numbers

b. Find four numbers in A.P whose sum is 20 and sum of squares is 120.

Answer:

a. The numbers are a – d, a, a + d

a-d + a + a + d = -3

Question 7.

a. The third term of a GP is 4. Find the product of its first 5 terms.

b. If the sum of the geometric series 1 + 4 + 16 +…. is 5461. Find the number of terms.

Answer:

Question 8.

a. Find the sum of the series

to 12 terms

b. Find the sum of the series 2+16+18+………..+4374

Answer:

Question 9.

Find the sum to n terms of the series 0.4 + 0.44 + 0.444 + …. to n terms

Answer:

Question 10.

The sum of three numbers in A.P is -3 and their product is 8. Find the numbers.

Answer:

The numbers are a – d, a, a + d

a-d +a+a + d = -3

3a = -3, a = -1

(a – d) (a – d) a = 8, (a^{2} – d^{2}) a = 8

1 – d^{2} = – 8, d^{2} = 9 d = ±3

The numbers are 2,-1, – 4

**Long Answer** **Type Questions **

**(Score 6)**

Question 1.

a. The inventor of the chess board suggested a reward of one grain of wheat for the first square, 2 grains for the second, 4 grains for the third and so on, doubling the number of the grains for subsequent squares. How many grains would have to be given to inventor (consider 64 squares in the chess boards).

b. The nth term of the GP,

Answer:

Question 2.

a The first, second and last terms of an A.P are a, b, c respectively. Show that the sum of the A.P is…..

b. If a^{2} + b^{2}, ab + be and b^{2} + c^{2} are in GP, Prove that a, b, c are in GP.

Answer:

Question 3.

a. If m times the m^{th} term of an A.P is equal to n times its n^{th} term. Show that the (m + n)^{th} term of the A.P is zero, h In a GP, which consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression is equal to …………

a. 1/2(1-)

b. 1/2

c.

d. 1/2 ()

Answer:

Question 4.

a. The product of first 3 terms of a GP is 1000. If 6 is added to the second term and 7 is added to the third term, the terms becomes an A.P.

i. Find the second term of GP

ii. Find the terms of the Gp.

iii. Find the sum of n terms of the series 7 + 77 + 777 + …………

Answer:

Question 5.

a. If the A.M between two positive numbers is 34 and their GM is 16. Find the numbers.

b. The sum to infinity of

Answer:

Question 6.

a Find the sum of n terms of the series whose n^{th} term is n (n + 3).

b. Find the n^{th} term of the series 1.3 + 3.5 + 5.7 +…………. . Also find the sum to n terms.

Answer:

Question 7.

a Sum the series 1.2.3+ 2.3.4+3.4.5+ …. to n terms.

b. If a b, c and d are in GP. Show that (b – c)^{2 }+ (c – a)^{2} + (d – b)^{2} = (a – d)^{2 }Answer:

Question 8.

The sequence N of natural numbers is divided into classes as follows.

Show that the sum of the numbers in the nth row is n (2n^{2} +1).

Answer:

Number of terms in first row = 2

Number of terms in second row = 4

Number of terms in third row = 6

…… …….. ……. …… ……. …….. …..

…… …….. ……. ……. ……. …….. ……

Number of terms in nth row = 2n

Now, number of terms upto the end of nth row = 2 + 4 + 6 + ….+ 2n

Also, number of terms upto the end of the (n-1 )^{th} row

=(n -1 )^{2} + (n – 1) = n^{2} – 2n + 1 + n -1 = n^{2} – n Sum of terms in nth row

∴ Sum of terms upto the end of nth row – Sum of terms upto the end of (n-1)^{th} row.

**NCERT Questions and Answers**

Question 1.

Write the first three terms in a_{n} = 2n + 5

Answer:

Substituting n = 1,2, 3, we get

a_{1} ^{=} 2(1) + 5 = 7, a_{2} = 9, a_{3} = 11

Therefore, the required terms are 7,9 and 11.

Question 2.

What is the 20^{th} term of the sequence defined by a_{n} = (n -1) (2 – n) (3 + n) ?

Answer:

‘Putting n = 20 , we obtain

^{a}_{20} ^{=} (20 – 1) (2 – 20) (3 + 20)

= 19 ×(-18)×(23) =-7866

Question 3.

The income of a person is Rs. 3,00,000, in the first year and he receives an increase of Rs.10,000 to his income per year for the next 19 years. Find the total amount, he received in 20 years.

Answer:

Here, we have an A.P. with a = 3,00,000, d = 10,000, and n=20.

Using the sum formula, we get,

S_{20}=

=10 ×(790000)=79,00,000.

Hence, the person received Rs. 79,00,000 as the total amount at the end of 20 years.

Question 4.

Insert 6 numbers between 3 and 24 such that the resulting sequence is an P.

Answer:

Here, a = 3, b = 24, n = 8.

Therefore, 24 = 3 + (8-1)d, so that d = 3.

Thus A_{1} = a + d = 3 + 3 = 6;

A_{2} = a + 2d = 3+2×3 = 9;

A_{3} = a + 3d = 3 + 3 × 3 = 12;

A_{4} = a + 4d = 3 + 4 × 3 = 15;

A_{5} = a + 5d = 3 + 5×3 = 18;

A_{6} = a + 6d = 3 + 6×3 = 21.

Hence, six numbers between 3 and 24 are 6,9, 12,15,18 and 21.

Question 5.

The ratio of the sums of m and n terms of an A.P is m^{2}: n^{1}. show that the ratio of m^{th} and n^{,th}terms is (2m -1): (2n -1)

Answer:

Question 6.

In a sequence n^{th} term is given by Find 20^{th} term.

Answer:

Question 7.

Find the sum to n terms of the series 5 + 55 + 555 + ………….

Answer:

Question 8.

If are in A.P .Prove that a,b,c, are in A.P.

Answer:

Question 9.

a. Find the 5^{th} term of the sequence whose n^{th}term,

b. Find 7 + 77 + 777 +7777 + …………….. to n terms.

Answer:

a.