Question

If the ap of the sum of first ten terms is -150 and the sum of next ten terms is -550 .find the ap.

Answer 1

Let the first  term of  AP be a  and common difference  be d  

Now,

Sum of  first  10 terms   = -150

$\frac{10}{2}(2a + (10-1)d) = -150\\\\5(2a+9d)=-150\\\\2a+9d= -30$

Again, given  that sum of  next  10 terms   is  -550, Therefore,

sum of  first  20 terms   =  -150  + (-550)  = -700

$\frac{20}{2}(2a + (20-1)d)=-700\\\\10(2a+ 19d)=-700\\\\2a+ 19d=-70\\\\\text{Subtracting both equations}\\\\-10d=40\\\\d=-4\\\\ \text{Putting d = -4 in 2a + 9d = -30 }\\\\2a-36=-30\\\\2a=-30+36\\ \\2a=6\\\\a=3$

a = 3

d = - 4

Answer 2

"Then the AP will be 3, -1, -5, -9,….

Given:

Sum of first ten terms of an AP = -150

Sum of next ten terms of an AP =-550

To find:

AP

Solution:

Sum of n terms of an AP $= \frac { n } { 2 } ( 2 a + ( n - 1 ) d )$

Sum of first 10 terms $= \frac { 10 } { 2 } ( 2 a + 9 d )$

-150 = 5(2a+9d)

-150=10a+45d…..(i)

Sum of next 10 terms $= \frac { 20 } { 2 } ( 2 a + 19 d )$

-150-550=10(2a+19d)

-700=10(2a+19d)

-70=2a+19d …….(ii)

Combining 1 and 2, we get

Multiplying (ii) by 5

-350=10a+95d….(iii)

Subtracting (i) from (iii) we get,

-350+150=10a+95d-10a-45d

-200=50d

d=-4

substitute d=-4 in equation (i) we get

-150=10a+45d

-150=10a+45(-4)

-150=10a-180

10a=-150+180

10a=30

a=3

hence, a=3 and d=-4

The AP will be a, a+d, a+2d, a+3d,…

Then the AP will be 3, -1, -5, -9,…."