Question

The 2nd and 45th term of an AP are 10 and 96 respectively.
Find the first term and the common difference and hence find the sum of the first 15 terms.

Answer 1

a+d=10

a+44d=96

Solve them simultaneously to get a and d . The use the formula S=n/2 (2a+n-1×d)

Answer 2

First term = 8, Common difference = 2, Sum of first 15 terms is 330

Step-by-step explanation:

Given Data

2nd term of AP = 10

45th term of AP = 96

Find the first term (a), common difference (d), and the sum of first 15 terms ($S_1_5$)

The Arithmetic Progression series can be written as

a, a+d, a+2d, a+3d, ...etc.,

First term is 'a' and second term is a+d

45th term can be find by using the formula

$a_n = a+(n-1)d$

$a_4_5 = a+ 44d$

Now a + d = 10 -------> (1)

       a + 44d = 96 ------> (2)

On solving the above equation we get,  

- 43d = - 86

Eliminate the negative sign on both sides

$d = \frac{86}{43}$

d = 2

Substitute d= 2 in equation (1)

a = 10 - 2 = 8

a = 8

Sum of first 15 terms can be calculated by using the formula

$S_1_5 = \frac{n}{2} [2a+(n-1) d]$

Substitute the respective values in above equation

$S_1_5 = \frac{15}{2} [16+ (15 - 1) (2)]$

$S_1_5 =\frac{15}{2} [16 + 14(2)]$

$S_1_5 =\frac{15}{2} [44]$

$S_1_5 = 15 \times 22$

$S_1_5 = 330$

Therefore the first term (a) is  8, the common difference (d) is 2 and the sum of first 15 series ($S_1_5$) is 330