Question

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Answer 1

Question:

The sum of first 10 terms of an A.P is -150 and the sum of its next 10 terms is -550. Find the A.P.

Answer:

The A.P is 3, -1, -5.......

Step-by-step explanation:

Given:

• Sum of first 10 terms = -150
• Sum of next 10 terms = -550

To Find:

• The A.P

Solution:

➜ We know that in an A.P, sum of n terms is given by,

    $\tt{S_n=\dfrac{n}{2}(2a_1+(n-1)\times d)}$

    where Sₙ = sum of n terms

    n = number of terms

    a₁ = first term

    d = common difference

➜ By given,

    Sum of first 10 terms = -150

➜ Hence,

    S₁₀ = 10/2 ( 2a₁ + (10 - 1) × d)

    5 (2a₁ + 9d) = -150

    2a₁ + 9d = -30---------(1)

➜ Also by given,

    Sum of next ten terms = -550

➜ Therefore

    Sum of first 20 terms = -150 + -550 = -700

    S₂₀ = 20/2 (2a₁ + (20 - 1) × d)

    10 (2a₁ + 19d) = -700

    2a₁ + 19d = -70--------(2)

➜ Solving equation 1 and 2 by elimination method.

    2a₁ + 19d = -70

    2a₁ + 9d = -30

             10d = -40

                 d = -40/10

                d = -4

➜ Hence common difference of the A.P is -4.

➜ Substituting the value of d in equation 1,

     2a₁ + 9 × -4 = -30

     2a₁ - 36 = -30

     2a₁ = 6

     a₁ = 3

➜ Hence first term of the A.P is 3.

➜ Now,

   Second term = a₁ + d = 3 + -4 = -1

   Third term = a₁ + 2d = 3 + -8 = -5

➜ Therefore the A.P is 3, -1, -5..........

Answer 2

Given

• Sum of 10 terms = -150
• Sum of next ten terms = -550

To find

• Required AP.

Solution

By using formula

$\boxed{S_n = \dfrac{n}{2}[2a + (n - 1)d]}$

Now, according to the question

→ $S_10 = \frac{10}{2}[2a + (10 - 1)d]$

→ $-150 = 5[2a + 9d]$

→ $-150 = 10a + 45d$

→ $10a + 45d + 150 = 0$

or

→ $2a + 9d = -30$⠀⠀⠀⠀.....(1)

Similarly,

→ $S_20 = \frac{20}{2}[2a + (20 - 1)d]$

→ $(-150) + (-550) = 10[2a + 19d]$

→ $700 = 20a + 190d$

→ $20a + 190d + 7000 = 0$

or

→ $2a + 19d = -70$⠀⠀⠀.....(2)

By elimination method

From (1) and (2) ,

2a + 19d = -70

2a + 9d = -30

⠀⠀-⠀⠀ ⠀ +

0a + 10d = -40

⠀⠀⠀10d = -40

⠀⠀⠀⠀⠀d = -4

=> Substituting the value of d in Equation (1)

2a + 9 × (-4) = -30

2a - 36 = -30

2a = 6

a = 3

Now,

★ Second term = a₁ + d

= 3 + -4

= -1

★ Third term = a₁ + 2d

= 3 + -8

= -5

Hence,

The required A.P = 3, -1, -5, ......