Question

If the sum of first 10 terms of an AP is -60 and sum of the first 15 terms is -165,then find the sum of its n terms.

Answer 1

Here is the answer
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Answer 2

The sum of an AP's n terms is Sₙ = 4n-n² if the sum of the first 10 terms is -60 and the sum of the first 15 terms is -165.

Given that,

We have to find the sum of an AP's n terms if the sum of the first 10 terms is -60 and the sum of the first 15 terms is -165.

We know that,

We know that,

Sum of the AP formula that is

Sₙ = $\frac{n}{2}$(2a+(n-1)d]

We have the sum of the first 10 terms i -60

-60 = $\frac{10}{2}$(2a+(10-1)d]

-60 =5 (2a+9b)
$\frac{-60}{5}$ = 2a+9b

2a+9b = -12     -------------------->equation(1)

Now,

We have the sum of the first 15 terms i -165

-165 = $\frac{15}{2}$(2a+(15-1)d]

-165 =$\frac{15}{2}$(2a+14b)

-165 = 15(a+7b)
$\frac{-165}{15}$ = a+7b

a+7b = -11    -------------------->equation(2)

Now by substituting equation (1) and (2)

We get,

                               2a+9b = -12

(a+7b = -11)×2⇒      2a+14b=-22

                            -------------------------(Subtracting)

                               0+-5b = 10

                                      b=$\frac{-10}{5}$

                                      b=-2

Substitute in equation(1)

2a+9b=-12

2a+9(-2)=-12

2a -18 =-12

2a = -12+18

2a = 6

a=3

So, The sum of n terms is

Sₙ = $\frac{n}{2}$(2a+(n-1)d]

Sₙ = $\frac{n}{2}$(2(3)+(n-1)(-2)]

Sₙ = $\frac{n}{2}$(6-2n+2]

Sₙ = $\frac{n}{2}$(-2n+8)

Sₙ = n(-n+4)

Sₙ = 4n-n²

Therefore, the sum of an AP's n terms is Sₙ = 4n-n² if the sum of the first 10 terms is -60 and the sum of the first 15 terms is -165.

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