Question

The first term of an AP is 10, the last term is 105 and the sum is 1150. Find the number of terms and the common difference.

Answer 1

$\blue{\mathbb{\underline{ANSWER:}}}$

• The number of terms = 20.

• Common difference = 5.

$\orange{\mathbb{\underline{GIVEN:}}}$

• The first term of an AP is 10

• The last term is 105

• Sum is 1150.

$\green{\mathbb{\underline{TO \ FIND:}}}$

• The number of terms.

• Common difference.

$\purple{\mathbb{\underline{EXPLANATION:}}}$

$\boxed{ \bold{S_n =\dfrac{n}{2}(a+l)}}$

$\tt{S_n= 1150}$

a = 10

l = 105

$\tt{1150 =\dfrac{n}{2}(10+ 105)}$

2300 = n(115)

n = 20

$\boxed{ \bold{n = \dfrac{l - a}{d} + 1 }}$

n = 20

l = 105

a = 10

$\sf{20 = \dfrac{105 - 10}{d}+1}$

$\sf{20 = \dfrac{95 + d}{d}}$

$\sf{20d =95 + d}$

$\sf{19d =95}$

$\sf{d =5}$

Hence the common difference is 5 and number of terms is 20.

$\red{\mathbb{\underline{VERIFICATION:}}}$

$\boxed{\bold{S_n = \dfrac{n}{2} (2a + (n-1)d)}}$

n = 20

a = 10

d = 5

$\sf{S_n = \dfrac{20}{2} (2(10) + (20-1)5)}$

$\sf{S_n = 10 (20+ 19(5))}$

$\sf {S_n = 10 (20+95)}$

$\sf{S_n = 10 (115)}$

$\sf{S_n = 1150}$

HENCE VERIFIED.

Answer 2

Given ,

First term (a) = 10

nth term (an) = 105

Sum of first n terms (Sn) = 1150

We know that , the sum of first n terms of an AP is given by

$\boxed{ \sf{S_{n} = \frac{n}{2} (a + a_{n} )}}$

Thus ,

1150 = n/2 × (10 + 105)

2300 = 115n

n = 2300/115

n = 20

Since , nth term (an) = 105

Thus ,

105 = 10 + (20 - 1)d

95 = 19d

d = 95/19

d = 5

Therefore ,

• The number of terms and the common difference of given AP are 20 and 5