Question

in an A.P. if 7th term is 91 and the sum of first four terms is 40. find the sum of first ten terms
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Answer 1

Step-by-step explanation:

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

• The 7th term of an AP is 91.

• The sum of first four terms is 40.

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

• The sum of first ten 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:-

• $\rm a_{n}$ = nth term

• a = first term

• n = number of terms

• d = common difference

The 7th term = 91

${ \rm \implies a_{7} = a + (7 - 1)d }$

${ \rm \implies a + 6d }= 91 .......(i)$

As we know that:-

Sum of n terms is given by the formula:-

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

Here:-

• $\rm S_{n}$ = sum of n terms

Sum of four terms = 40

${ \rm \implies S_{4} = \dfrac{4}{2} \bigg \{ 2a + (4 - 1)d \bigg \} }$

${ \rm \implies 40 = 2( 2a + 3d) }$

${ \rm \implies \dfrac{40}{2} = ( 2a + 3d) }$

$\rm \implies 20 = 2a + 3d .....(ii)$

Multiplying equation (ii) with 2

→ 2(2a + 3d = 20)

→ 4a + 6d = 40......(iii)

Subtracting equation (i) from (iii)

→ 4a + 6d - ( a + 6d ) = 40 - 91

→ 4a + 6d - a - 6d = -51

→ 3a = -51

→ a = $\dfrac{ - 51}{3}$

→ a = -17

Substituting a = -17 in equation (i)

→ a + 6d = 91

→ -17 + 6d = 91

→ 6d = 91 + 17

→ 6d = 108

→ d = 18

We have a = -17 and d = 18

The sum of first 10 terms

$\rm= \dfrac{n}{2} \bigg \{2a + (n - 1)d \bigg \}$

$\rm= \dfrac{10}{2} \bigg \{2 \times - 17+ (10 - 1)18\bigg \}$

$\rm= 5 \bigg \{(2 \times - 17)+ (9 \times 18)\bigg \}$

$\rm= 5 \bigg \{ - 34+ 162\bigg \}$

= 5 × 128

= 640

Therefore; The sum of first 10 terms is 640