Question

In the AP. 7, 14,21. .... How many terms are to be
Considered for getting
Sum 57402​

Answer 1

Question:

In the A.P 7, 14, 21, .... , How many terms are to be considered for getting the sum of 5740 ?

Answer:

$\bigstar{\bold{Number\:of\:terms=40}}$

Step-by-step explanation:

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

• The A.P is 7, 14, 21.....
• Sum of the A.P ($S_n$)= 5740

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

• Number of terms (n)

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

→ Here we have to find the number of terms(n) required to get a sum of 5740.

→ First we have to find the common difference (d) of the A.P

  d = a₂ - a₁

 Here a₂ = 14, a₁ = 7

→ Hence

  d = 14 - 7

  d = 7

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

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

→ Substituting the datas we get,

  $\sf{5740=\dfrac{n}{2}(2\times 7 +(n-1)\times 7)}$

  $\sf {5740=\dfrac{n}{2}(14+7n-7)}$

  $\sf {5740\times 2=n (7n+7)}$

  $\sf{7n^{2}+7n = 11480}$

→ Solving by factorization metod,

  $\sf {x=\dfrac{-b\pm \sqrt{b^{2}-4ac } }{2a}}$

 where a = 7, b = 7 , c= -11480

→ Substituting the datas,

  $\sf {x=\dfrac{-7\pm \sqrt{49-4\times 7\times -11480} }{2\times 7} }$

  $\sf {x=\dfrac{-7\pm \sqrt{49+321440} }{14}}$

  $\sf {x=\dfrac{-7\pm 567}{14}}$

→ Case 1:

  x = ( -7 - 567)/14

  x = -574/14

  x = -41

→ This cannot happen since number of terms can't be negative.

→ Case 2:

   x = (-7 + 567)/14

   x = 560/14

   x = 40

→ Hence the number of terms required for getting a sum of 5740 is 40

$\boxed{\bold{Number\:of\:terms=40}}$

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

→ Sum of  terms of an A.P is given by the formula,

   $\sf {S_n=\dfrac{n}{2} (a_1+a_n )}$

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