Question

first term=5&nth term is 45.sum is 400,find the no. of terms and the d​

Answer 1

$\small\bold{first \: term \: a=5} \\ \bold{last \: term \: l=45 \: and \: S_n = 400}$

$\bold{Since, S_n = \frac{n}{2} (a + 1)}$

$\bold{ \implies400 = \frac{n}{2} (5 + 45)}$

$\bold{ \implies400 = \frac{n}{2} (50)}$

$\bold{\implies400 = n(25)}$

$\bold{ \implies \: n = 16}$

Since, l=a+(n−1)d

⇒45=5+(16−1)d

⇒45−5=(15)d

⇒40=15d

$\implies \: d = \frac{40}{15}$

$⇒d = \frac{8}{3}$

Answer 2

Step-by-step explanation:

${\Large{\bf{\underline{\purple{Question :}}}}}$

First term of AP is 5 and nth term is 45. Sum is 400, find the no. of terms and the d

${\Large{\bf{\underline{\purple{Solution :}}}}}$

Given :

• First term (a) = 5
• Last term (l) = 45
• Sum = 400

Since the last term is given, we can use the formula

$\\ {\large{\red{✠}}} \: \: \: \: \: {\large{\boxed{\sf{S_n \: = \: \frac{n}{2} \: (a \: + \: l)}}}}$

Putting values in formula

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: 400 \: = \: \frac{n}{2} (5 \: + \: 45)}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: 400 \: = \: \frac{n}{2} × 50}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: \frac{400 \: × \: 2}{50} \: = \: n}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: 8 \: \times \: 2 \: = \: n}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: 16 \: = \: n}}}$

${\large{\sf{\fbox{\red{So, number of terms \: = \: 16}}}}}$

Finding d,

For d we use the formula

${\large{\red{✠}}} \: \: \: \: \: {\large{\boxed{\sf{S_n \: = \: \frac{n}{2} \: [(2a \: + \: n \: - \: 1) d]}}}}$

${\large{\mathrm{\green{Putting \: \: \: S_n \: = \: 400, \: \: n \: = \: 16, \: \: a \: = \: 5}}}}$

${\large{\purple{⟹ \: \: \: \: \: \: \: \: \: \: }}}{\bold{\sf{400 \: = \: \frac{16}{2} \: [(2 \: × \: 5 \: + \: 16 \: - \: 1) \: × \: d]}}}$

${\large{\purple{⟹ \: \: \: \: \: \: \: \: \: \: }}}{\bold{\sf{400 \: = \: 8 \: (10 \: + \: 15d)}}}$

${\large{\purple{⟹ \: \: \: \: \: \: \: \: \: \: }}}{\bold{\sf{\frac{400}{8} \: = \: 10 \: + \: 15d}}}$

${\large{\purple{⟹ \: \: \: \: \: \: \: \: \: \: }}}{\bold{\sf{50 \: = \: 10 \: + \: 15d}}}$

${\large{\purple{⟹ \: \: \: \: \: \: \: \: \: \: }}}{\bold{\sf{50 \: - \: 10 \: = \: 15d}}}$

${\large{\purple{⟹ \: \: \: \: \: \: \: \: \: \: }}}{\bold{\sf{40 \: = \: 15d}}}$

${\large{\purple{⟹ \: \: \: \: \: \: \: \: \: \: }}}{\bold{\sf{\frac{40}{15} \: = \:d}}}$

${\large{\purple{⟹ \: \: \: \: \: \: \: \: \: \: }}}{\bold{\sf{\frac{8}{3} \: = \:d}}}$

${\large{\boxed{\mathrm{\red{So, common \: difference \: = \: \frac{8}{3}}}}}}$