Question

if an A.P. sum of first 10 terms is -150 and the sum of next ten terms is -550. Find the A.P.​

Answer 1

Answer:

$\bigstar{\bold{The\:A.P\:is\:3,-1,-5....}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

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

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• The A.P

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ The sum of n terms of an A.P is given by the formula,

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

→ By given the sum of first 10 terms of the A.P is given by

  $\sf{S_{10}=\dfrac{10}{2}(2a_1+(10-1)\times d)}$

  $\sf{S_{10}=5(2a_1+9d)}$

  $\sf{-150=5(2a_1+9d)}$

  $\sf{10a_1+45d=-150---(1)}$

→ Also by given the sum of next 10 terms = -550

→ Hence,

  Sum of 20 terms  = Sum of first ten terms +  Sum of next ten terms

  Sum of 20 terms = -150 + -550

  Sum of 20 terms = -700

→ Hence,

  $\sf{S_{20}=\dfrac{20}{2}(2a_1+(20-1)\times d)}$

  $\sf{-700=10(2a_1+19d)}$

  $\sf{20a_1+190d=-700---(2)}$

→ Multiply equation 1 by 2

  20a₁ + 90d = -300---(3)

→ Solve equation 2 and 3 by elimination method,

  20a₁ + 190d = -700

  20a₁ + 90d = -300

             100d = -400

                   d= - 400/100

                   d = -4

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

→ Substitute the value of d in equation 2

  20a₁ + 190 × -4 = -700

  20a₁ - 760 = -700

  20a₁ = 60

       a₁ = 3

→ Hence first term of the A.P is 3

→ Second term a₂ = a₁ + d

  Second term = 3 + -4 = -1

→ Third term a₃ = a₂ + d

  Third term = -1 + -4 = -5

→ Hence the A.P is 3, -1, -5.....

$\boxed{\bold{The\:A.P\:is\:3,-1,-5....}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ Sum of n terms of an A.P is given by

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

  $\sf{S_n=\dfrac{n}{2}(a_1+a_n)}$