Question

In an AP the sum of first 10 terms is -150 and the sum of next ten terms is -550.Find the AP?

class 1 question​

Answer 1

Answer:

3,-1,-5...

Step-by-step explanation:

Note: Sum of first n terms Sn = (n/2)[2a + (n - 1) * d]

(i)

Given, Sum of first 10 terms is -150.

⇒ S₁₀ = (10/2)[2a + (n - 1) * d]

⇒ -150 = 5[2a + (10 - 1) * d]

⇒ -30 = 2a + 9d

(ii)

Given, Sum of next 10 terms is --550.

∴ S₂₀ - S₁₀ = -550

⇒ S₂₀ = -550 - 150

⇒ S₂₀ = -700

Now,

⇒ S₂₀ = (20/2)[2a + (20 - 1) * d]

⇒ -700 = 10[2a + 19d]

⇒ -70 = 2a + 19d

On solving (i) & (ii), we get

⇒ 2a + 9d = -30

⇒ 2a + 19d = -70

   --------------------

            -10d = 40

                d = -4.

Substitute d = -4 in (i), we get

⇒ 2a + 9d = -30

⇒ 2a + 9(-4) = -30

⇒ 2a - 36 = -30

⇒ 2a = -30 + 36

⇒ a = 3.

Therefore, the AP is 3,-1,-5...

Hope it helps!

Answer 2

$\huge \bf \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}$

$\bf{QUESTION}$

In an AP the sum of first 10 terms is -150 and the sum of next ten terms is -550.Find the AP?

$\bf{step \: by \: step \: explanation}$

Given :-

=> sum of first ten terms of an AP = -150

=> sum of next ten terms of an AP = -550

°•° Sum of first n terms can be written as

Sn = (n/2)[2a + (n - 1) * d]

Case 1 .
•••••••••••

→ S10 = (10/2)[2a + (n - 1) × d]

→ -150 = 5[2a + (10 - 1) × d]

→ 2a + 9d = -30.....( 1 )

Case 2
•••••••••••

S20 - S10 = -550

→ S20 = -550 - 150

→ S20 = -700

°•° S20 = (20/2)[2a + (20 - 1) × d]

→ -700 = 10[2a + 20d - d ]

→ -700 = 10[2a + 19d]

→ 2a + 19d = -70......(2)

from equation ( 1 ) & ( 2 )

2a + 9d = -30
2a + 19d = -70
____________
  -10d = 40

$\bf \: { \implies \: d \: = - 4}$

put the value of d in equation 1 .

→ 2a + 9d = -30

→ 2a + 9(-4) = -30

→ 2a - 36 = -30

→ 2a = -30 + 36

→ 2a = 6

→ a = 6/3

$\bf{ \implies \: a = 3}$

Hence ,

=> a = 3

=> a + d = 3 + -4

→ -1

=> a + 2d = 3+ 2×-4

=> 3 + (-8)

→ -5

•°• AP is 3,-1,-5........

$\huge \pink{ \boxed{ \boxed{ \mid{ \ulcorner{ \mathbb{THANKS}}}}}}$