How many terms are there in an A.P. whose fir | vedclass

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(C) Given: First term $a = -14$ and fifth term $a_5 = 2$.Let $d$ be the common difference of the $A.P.$Using the formula $a_n = a + (n-1)d$,we have $a

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$10$

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(C) Given: First term $a = -14$ and fifth term $a_5 = 2$.Let $d$ be the common difference of the $A.P.$Using the formula $a_n = a + (n-1)d$,we have $a_5 = a + 4d = 2$.Substituting $a = -14$: $-14 + 4d = 2 \Rightarrow 4d = 16 \Rightarrow d = 4$.Let $n$ be the number of terms such that the sum $S_n = 40$.Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:$40 = \frac{n}{2}[2(-14) + (n-1)4]$$80 = n[-28 + 4n - 4]$$80 = n[4n - 32]$$80 = 4n^2 - 32n$Dividing by $4$: $n^2 - 8n - 20 = 0$.Factoring the quadratic equation: $(n - 10)(n + 2) = 0$.Thus,$n = 10$ or $n = -2$.Since the number of terms $n$ cannot be negative,$n = 10$.

How many terms are there in an $A.P.$ whose first and fifth terms are $-14$ and $2,$ respectively,and the sum of terms is $40$?

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