Question

in an ap the sum of the first 10th terms is -150and the sum of the 10th term is -550 .find the ap

Answer 1

Dear student,

Solution:

Sum of n elements Sn = $\frac{n}{2}( 2a+ (n-1)d)$

Sum of first 10 elements S10 = $\frac{10}{2}( 2a+ (10-1)d)$

$\frac{10}{2}( 2a+ (10-1)d)$ = -150

5(2a+9d) = - 150

2a+9d = - 30 -------eq1

10 th term is -550

Tn = a+(n-1)d

T10 = a+9d = -550

a+9d = -550 -------eq2

Subtract both equation 1 and 2

2a+9d-a-9d = - 30 + 550

a = 520

a+9d = -550

520 +9d = -550

9d = -550 -520

9d = -1070

d = -1070/9

d = -118.8