Question

in an AP the first term is 22 and term is minus 11 and the sum of first n terms is 66 find and and the common difference​

Answer 1

Answer:

Appropriate Question :-

• In an AP the first term is 22, nth term is - 11 and the sum of first n terms is 66. Find the number of terms n and the common difference d.

Given :

• In an AP the first term is 22, nth term is - 11 and the first n terms is 66.

To Find :-

• What is the number of terms n and the common difference d.

Formula Used :-

$\clubsuit$ Sum of Arithmetic Series of an AP Formula :

$\mapsto \sf\boxed{\bold{\pink{S_n =\: \dfrac{n}{2}\bigg(a + l\bigg)}}}$

where,

• $\sf S_n$ = Sum of nth terms
• n = Number of terms of an AP
• a = First term of an AP
• l = Last term or nth term of an AP

$\clubsuit$ General term (nth term) of an AP Formula :

$\mapsto \sf\boxed{\bold{\pink{a_n =\: a + (n - 1)d}}}$

where,

• $\sf a_n$ = nth term of an AP
• a = First term of an AP
• n = Number of terms of an AP
• d = Common difference of an AP

Solution :-

First, we have to find the number of terms (n) :

Given :

• Sum of first nth term ($\sf S_n$) = 66
• First term (a) = 22
• Last term or nth term (l) = - 11

According to the question by using the formula we get,

$\longrightarrow \sf 66 =\: \dfrac{n}{2}\bigg(22 + \{- 11\}\bigg)$

$\longrightarrow \sf 66 =\: \dfrac{n}{2}\bigg(22 - 11\bigg)$

$\longrightarrow \sf 66 =\: \dfrac{n}{2}\bigg(11\bigg)$

$\longrightarrow \sf 66 =\: \dfrac{n}{2} \times 11$

$\longrightarrow \sf 66 =\: \dfrac{11n}{2}$

$\longrightarrow \sf 11n =\: 2(66)$

$\longrightarrow \sf 11n =\: 2 \times 66$

$\longrightarrow \sf 11n =\: 132$

$\longrightarrow \sf n =\: \dfrac{\cancel{132}}{\cancel{11}}$

$\longrightarrow \sf n =\: \dfrac{12}{1}$

$\longrightarrow \sf\bold{\red{n =\: 12}}$

${\small{\bold{\underline{\therefore\: The\: number\: of\: terms\: of\: an\: AP\: is\: 12\: .}}}}$

Now, we have to find the common difference (d) :

Given :

• nth term of an AP ($\sf a_n$)= - 11
• First term of an AP (a) = 22
• Number of terms of an AP (n) = 12

According to the question by using the formula we get,

$\longrightarrow \sf - 11 =\: 22 + (12 - 1)d$

$\longrightarrow \sf - 11 =\: 22 + (11)d$

$\longrightarrow \sf - 11 =\: 22 + 11d$

$\longrightarrow \sf - 11 - 22 =\: 11d$

$\longrightarrow \sf - 33 =\: 11d$

$\longrightarrow \sf \dfrac{- \cancel{33}}{\cancel{11}} =\: d$

$\longrightarrow \sf \dfrac{- 3}{1} =\: d$

$\longrightarrow \sf - 3 =\: d$

$\longrightarrow \sf\bold{\red{d =\: - 3}}$

${\small{\bold{\underline{\therefore\: The\: common\: difference\: of\: an\: AP\: is\: -\: 3\: .}}}}$

Answer 2

$\large \sf \underline\bold \red {Given : - }$

First term, (a) = 22

nth term, (l) = -11

Sum of nth term, (Sn) = 66

$\large \sf \underline\bold \pink{Explanation\::}$

Using formula of sum of an A.P;

$\boxed{\bf{S_n = \frac{n}{2} [a+l]}}$

A/q

$\mapsto\tt{S_n = \dfrac{n}{2} \times [a+l]}$

$\mapsto\tt{66 = \dfrac{n}{2} \times [22+(-11)]}$

$\mapsto\tt{66 = \dfrac{n}{2} \times [22-11]}$

$\mapsto\tt{66 = \dfrac{n}{2} \times 11}$

$\mapsto\tt{66 \times 2 = 11n}$

$\mapsto\tt{132 = 11n}$

$\mapsto\tt{n = \cancel{132/11}}$

$\mapsto\bf{n = 12}$

Now, using formula of the A.P;

$\boxed{\bf{a_n = a+(n-1)d}}$

$\mapsto\tt{-11 = 22 + (12- 1)d}$

$\mapsto\tt{-11 = 22 + (11)d}$

$\mapsto\tt{-11-22 = 11d}$

$\mapsto\tt{-33= 11d}$

$\mapsto\tt{d= -\cancel{33/11}}$

$\mapsto\bf{d = -3}$

Thus,

The A.P of  n & d will be 12 & -3 .