Question

• Third term of an arithmetic sequence is 8 and
its 5th term is 14. what is the sum of the first 9 terms?​

Answer 1

$\rm \green{\mapsto \:{ \underline{ \underline{given}}}}$

=> 3rd term of an Ap = 8

=> 5th term of an Ap = 14

$\rm \red{\implies{ \underline{to \: find}}}$

sum of the first 9 terms of an Ap

$\rm \blue{{ \underline{solution \div }}}$

Formula of Nth term of an Ap

=> An = a + ( n - 1 ) d

=> 3rd term = 8

=> a + 2d = 8 ......(1)

=> 5th term = 14

=> a + 4d = 14 ......(2)

From equation (1) and (2) we get,

=> -2d = -6

=> d = 3

put the value of d = 3 in equation (1)

we get,

=> a + 2(3) = 8

=> a + 6 = 8

=> a = 2

Therefore,

=> First term = a = 2

=> common difference = d = 3

Formula of sum of Nth term of an Ap

$\green{\rm \to \: s_{n} \: = \frac{n}{2}(2a \: + \: (n - 1)d) }$

$\rm \to \: s_{9} \: = \frac{9}{2}(2 \times 2 \: + \: 8 \times 3)$

$\rm \to \: s_{9} \: = \frac{9}{2}(4 + 24)$

$\rm \to \: s_{9} \: = \frac{9}{2}(28)$

$\rm \to \: s_{9} \: = \:9 \times 14$

$\rm \to \: s_{9} \: = 126$

Answer 2

Step-by-step explanation:

$\bf{\underline{\underline\red{Given:-}}}$

• Third term of an AP is 8

• And the fifth term of the AP is 14.

$\bf{\underline{\underline\blue{To\:Find:-}}}$

• The sum of first 9 terms.

$\bf{\underline{\underline\green{Solution:-}}}$

As we know that

The nth term of an AP is given by the formula:-

$\boxed{ \rm \leadsto a_{n} = a + (n - 1)d}$

Here:-

• a = first term

• n = number of terms

• d = common difference

The third term = 8

${ \rm \implies a_{3} = 8}$

${ \rm \implies a + (3 - 1)d = 8}$

${ \rm \implies a + 2d = 8......(i)}$

The fifth term = 14

${ \rm \implies a_{5} = 14}$

${ \rm \implies a + (5 - 1)d = 14}$

${ \rm \implies a + 4d = 14......(ii)}$

Subtracting equation (i) from (ii)

${ \rm \implies a + 4d - (a + 2d) = 14 - 8}$

${ \rm \implies a + 4d - a - 2d = 6}$

${ \rm \implies 4d - 2d = 6}$

${ \rm \implies 2d= 6}$

${ \rm \implies d= 3}$

Substituting d = 3 in equation (i)

${ \rm \implies a + 2d= 8}$

${ \rm \implies a + 2(3)= 8}$

${ \rm \implies a +6= 8}$

${ \rm \implies a = 8 - 6}$

${ \rm \implies a = 2}$

Now:- We have a = 2 and d = 3 and n = 9

Sum of first 9 terms:-

The sum of n terms is given by the formula

$\boxed{ \rm \leadsto S_{n} = \frac{n}{2} \bigg \{2a + (n - 1)d \bigg \}}$

Substituting the values

${ \rm \implies S_{9} = \dfrac{9}{2} \bigg \{2 \times 2 + (9 - 1)3 \bigg \}}$

${ \rm \implies S_{9} = \dfrac{9}{2} \bigg \{4 + (8)3 \bigg \}}$

${ \rm \implies S_{9} = \dfrac{9}{2} \bigg \{4 + 24\bigg \}}$

${ \rm \implies S_{9} = \dfrac{9}{2} \times 28}$

${ \rm \implies S_{9} = 9 \times 14}$

${ \rm \implies S_{9} = 126}$

$\large \underline{ \boxed{\rm \purple{ \therefore Sum \: of \: 9 \: terms = 126}}}$