Question

In an AP. the sum of first ten terms is -150 and the sum of its next ten terms is -530
Find the AP​

Answer 1

Given the sum of the first 10 terms of an AP is -150 and the sum of the next ten terms which means Sum of first 20 terms - Sum of first 10 terms is -530.

Let the first term, common difference of the AP be a and d respectively. Now,

⇒ S₁₀ = 10/2 [ 2a + (10 - 1)d ]

⇒ -150 = 5 [ 2a + 9d ]

⇒ 2a + 9d = -30 ...(i)

Now, according to the second case,

⇒ S₂₀ - S₁₀ = -530

⇒ S₂₀ - (-150) = -530

⇒ S₂₀ + 150 = -530

⇒ S₂₀ = -680

⇒ 20/2 [ 2a + (20 - 1)d ] = -680

⇒ 10 [ 2a + 19d ] = -680

⇒ 2a + 19d = -68 ...(ii)

Subtract (ii) from (i) , we get

⇒ 2a + 9d - (2a + 19d) = -30 - (-68)

⇒ 2a + 9d - 2a - 19d = -30 + 68

⇒ -10d = 38

⇒ d = -38/10

Now, substitute the value of d in (i), we get

⇒ 2a + 9×-38/10 = -30

⇒ 2a = -30 + 342/10

⇒ 2a = (-300 + 342) / 10

⇒ a = 42/10×2

⇒ a = 21/10

We know, the standard form of an AP is given by,

• a , a + d, a + 2d, a + 3d , ...

Substituting values ,we get

⇒ 21/10 , (-38+21)/10 , (21 - 76)/10 , (21 - 114)/10 , ...

⇒ 21/10 , -17/10 , -55/10 , -93/10 , ...

is the required AP.

Answer 2

Given :-

In an AP. the sum of first ten terms is -150 and the sum of its next ten terms is -530

To Find :-

AP

Solution :-

We know that

$\sf S_n = \dfrac{n}{2}\bigg(2a+(n-1)d\bigg)$

$\sf S_{10} = \dfrac{10}{2}\bigg(2a + (10-1)d\bigg)$

$\sf S_{10} = 5\bigg(2a+(10-1)d\bigg)$

$\sf S_{10} = 5\bigg(2a +9d\bigg)$

Now,

Putting S10 as -150

$\sf -150 = 5(2a+9d)$

$\sf \dfrac{-150}{5} = 2a+9d$

$\sf -30 =2a+9d$

Second case

$\sf S_{20} -( -150) = -530$

$\sf S_{20} + 150 = -530$

$\sf S_{20} = -530+(-150)$

$\sf S_{20} = -680$

Again by using the same formula

$\sf -680 = \dfrac{20}{2} \bigg(2a + (20-1)d\bigg)$

$\sf -680 = 10\bigg(2a+(20-1)d\bigg)$

$\sf \dfrac{-680}{10}=2a+19d$

$\sf -68 = 2a+19d$

Now,

$\sf 2a + 9d - (2a + 19d) = -30 - (-68)$

$\sf 2a + 9d -2a-19d = -30-(-38)$

$\sf 9d - 19d = -30+68$

$\sf -10d=38$

$\sf d = \dfrac{-38}{10}$

From 1 we get

$\sf 2a + 9\times \dfrac{-38}{10} =-30$

$\sf 2a + 9 \times -38=-30\times10$

$\sf 2a - 342=300$

$\sf 2a = 342-300$

$\sf a = \dfrac{21}{10}$

AP = 21/10 , -17/10 , -55/10 , -93/10 ,