Question

When first term ( a ) is 5 and their common difference is - 1 / 2 then find the sum of 16th term of an AP.

Answer 1

$\underline{\underline{\large{ \mathfrak{Solution : }}}}$



$\textsf{Given : } \\ \\ \mathsf{\implies First \: term (a) \: = \: 5 } \\ \\ \mathsf{\implies Common \: difference (d) \: = \: \dfrac{-1 \: }{\: 2 }} \\ \\ \mathsf{ \implies No. \: of \: terms \: = \: 16 }$



$\underline{\mathsf{To \: Find \: \rightarrow \: Sum \: of \: 16 \: terms. }}$



$\textsf{Using Formula : } \\ \\ \boxed{\mathsf{\implies S_{n} \: = \: \dfrac{n}{2}[2a \: + \: (n \: - \: 1)d ]}}$




$\underline{\textsf{Now,}} \\ \\ \mathsf{\implies S_{16} \: = \: \dfrac{16}{2}[2 \: \times \: 5 \: + \: ( 16 \: - \: 1)\left(\dfrac{-1}{2}\right)]}$



$\mathsf{ \implies S_{16} \: = \: 8 \left\{10 \: + \: 15 \left( \dfrac{ - 1}{2} \right) \right\} } \\ \\ \mathsf{ \implies S_{16} \: = \: 8 \left\{ 10 \: - \: \dfrac{15}{2}\right \}} \\ \\ \mathsf{ \implies S_{16} \: = \: 8 \left \{ \dfrac{20 \: - \: 15}{2} \right \}} \\ \\ \mathsf{ \implies S_{16} \: = \: 8 \: \times \: \dfrac{5}{2}}$


$\mathsf{\implies S_{16} \: = \: 4 \: \times \: 5 }\\ \\ \mathsf{\therefore \quad S_{16} \: = \: 20}$



$\boxed{\mathsf{Hence, \: sum \: of \: 16 \: terms \: of \: this \: A.P \: is \: 20.}}$

Answer 2

$According \: to \: question, \\ First \: term(a) = 5 \\ Common \: difference(d) = \frac{ - 1}{2} \\ \\ Sum \: of \: n \: terms \: of \: AP \\ = \frac{n}{2} \times[ 2a + (n - 1)d ]\\ \\ sum \: of \: 16 \: terms \: of \: AP \: \\ = \frac{16}{2} \times[ 2 \times 5 + (16 - 1)( \frac{ - 1}{2} ) ]\\ = 8 \times [10 - 7.5 ] \\ = 8 \times 2.5 \\ = 20 \\ \\ hope \: it \: may \: help \: u$