Question

how many terms are there in a.p whose first and fifth terms are -14 and 2 respectively and the sums of the terms is 40

Answer 1

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

• First term = -14
• fifth term = 2
• sum of terms = 40

$\large\bf\underline {To \: find:-}$

• Number of terms.

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

• First term (a) = -14

fifth term ( a+4d ) = 2

substituting value of a .

»» -14 +4d = 2

»» 4d = 2 + 14

»» 4d = 16

»» d = 4

So we get, common difference (d) = 4.

$\large\bf \: S_n = \frac{n}{2} \{2a + (n - 1)d \}$

$\dashrightarrow \rm\:40 = \frac{n}{2} \{ - 14 \times 2 + (n - 1)4 \} \\ \\ \dashrightarrow \rm\:40 = \frac{n}{2} \{ - 28 + 4n - 4 \} \\ \\ \dashrightarrow \rm\:40 \times 2 = n( - 32 + 4n) \\ \\ \dashrightarrow \rm80 = - 32n + 4 {n}^{2}$

On Divide both side by 4 we get,

$\dashrightarrow \rm20 = - 8n + {n}^{2} \\ \\ \dashrightarrow \rm {n}^{2} - 8n - 20 = 0 \\ \\ \dashrightarrow \rm {n}^{2} + 2n - 10n - 20 \\ \\ \dashrightarrow \rm \: n(n + 2) - 10(n + 2) \\ \\ \dashrightarrow \rm(n - 10)(n + 2) \\ \\ \dashrightarrow \rm \: n = 10 \: or \: n \: = - 2$

number of terms cannot be negative.

• Number of term = 10
• So, there are 10 term in AP.