Question

If the sum of the first 14 terms of an AP is 1050 and its first term is 10.
find the 20th term.
Solution : Hers
15​

Answer 1

$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\sf{sum \: of \: first \: 14 \: terms \: is \: 1050} \\ &\sf{first \: term \: is \: 10} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf To \: Find :- \begin{cases} &\sf{ {20}^{th} \: term \: of \: ap} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf Formula \: Used- \begin{cases} &\sf{S_n \: = \dfrac{n}{2}(2a \: + (n - 1)d) } \\ &\sf{a_n = a + (n - 1)d} \end{cases}\end{gathered}\end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

• Sₙ is sum of first n term.

Solution :-

$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\sf{a = 10} \\ &\sf{n = 14}\\ &\sf{S_{14} = 1050} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf Let :- \begin{cases} &\sf{common \: difference \: be \: d} \end{cases}\end{gathered}\end{gathered}$

$\tt \:  ⟼S_{14} = 1050$

$\tt\implies \:7\dfrac{ \cancel{14}}{\cancel2} ( 2 \times 10 + (14 - 1) \times d) = 1050$

$\tt\implies \:20 + 13d = \dfrac{1050}{7}$

$\tt\implies \:20 + 13d = 150$

$\tt\implies \:13d = 130$

$\large \boxed{\tt\implies \:d \: = 10}$

$\tt \:  ⟼Now, \: a_{20} \: = a \: + (20 - 1)d$

$\tt \:  ⟼ a_{20} = 10 + 19 \times 10$

$\tt \:  ⟼ \: a_{20} \: = 10 + 190$

$\large \boxed{\tt \:  ⟼ \: a_{20} = 200}$