Question

The sum of first 10 term of an AP is one third of the sum of next 10 terms .If first term is -5,find the sum of its first 30 terms.

Answer 1

Here's your answer

Isn't it simple!!!

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Answer 2

Answer:

-4500

Step-by-step explanation:

First term = a = -5

Formula of sum of first n terms = $S_n=\frac{n}{2}(2a+(n-1)d)$

Substitute n = 10

$S_{10}=\frac{10}{2}(2(-5)+(10-1)d)$

$S_{10}=5(-10+9d)$

Substitute n =20

$S_{20}=\frac{20}{2}(2(-5)+(20-1)d)$

$S_{20}=10(-10+19d)$

Now we are given that The sum of first 10 term of an AP is one third of the sum of next 10 terms

So, $S_{10}=\frac{1}{3}(S_{20}-S_{10})$

$5(-10+9d)=\frac{1}{3}((10(-10+19d))-(5(-10+9d)))$

$-50+45d=\frac{1}{3}(-100+190d-(-50+45d))$

$-50+45d=\frac{1}{3}(-100+190d+50-45d)$

$-150+135d=-50+145d$

$-150+50=145d-135d$

$-100=10d$

$-10=d$

So, sum of first 30 terms :

$S_{30}=\frac{30}{2}(2(-5)+(30-1)(-10))$

$S_{30}=15(-10-290)$

$S_{30}=-4500$

Hence the sum of first 30 terms is -4500