Question

The first term of an APis –5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.

Answer 1

hope the solution is useful....

Answer 2

Answer:

$\bigstar{\bold{Number\:of\:terms=6}}$

$\bigstar{\bold{Common\:difference=10}}$

Step-by-step explanation:

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

• First term (a₁) = -5
• Last term ($a_n$) = 45
• Sum of the AP ($S_n$) = 120

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

• Number of terms of the AP (n)
• Common difference (d)

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

Number of terms:

➤ The sum of n terms of an AP is given by the formula

    $S_n=\dfrac{n}{2} (a_1+a_n)$

➤ Substituting the given datas we get,

    120 = n/2 ( -5 + 45)

    n =120/20

    n = 6

$\boxed{\bold{Number\:of\:terms=6}}$

Common difference:

➳ Common difference of an AP is given by the formula

    $d=\dfrac{a_m-a_n}{m-n}$

    where $a_m$ = 45 , $a_n$ = -5 , m = 6, n =1

➳ Substituting the datas we get,

   $d=\dfrac{45+5}{6-1}$

➳ d = 10

$\boxed{\bold{Common\:difference=10}}$

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

➻ The common differennce of an AP is the difference between its two consecutive terms.

$d = a_2-a_1$

$d = \dfrac{a_m-a_n}{m-n}$

➻ Sum of n terms of an AP is given by

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

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