In an A.P. if 7th term is 91 and the sum of first four terms is 40. fi

1.	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
Answer» 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