the first term of an AP is 5, the last term is 45 and sum is 400,find the number of terms and the common difference
Answers
EXPLANATION.
=> First term of an Ap = 5
=> Last term of an Ap = 45
=> The sum = 400
To find number of terms and common
difference.
\begin{gathered}\sf \to \: as \: we \: know \: that \\ \\ \sf \to \: s_{n} = \dfrac{n}{2} (a + l) \\ \\ \sf \to \: put \: the \: value \: in \: equation \\ \\ \sf \to \: 400 = \frac{n}{2} (5 + 45) \\ \\ \sf \to \: 400 = 25n \\ \\ \sf \to \: n \: = 16\end{gathered}
→asweknowthat
→s
n
=
2
n
(a+l)
→putthevalueinequation
→400=
2
n
(5+45)
→400=25n
→n=16
\begin{gathered}\sf \to \: { \underline{some \: related \: formula}} \\ \\ \sf \to \: 1) = \: n \: terms \: of \: an \: ap \\ \\ \sf \to \: a_{n} = a + (n - 1)d \\ \\ \sf \to \: 2) = sum \: of \: nth \: terms \: of \: an \: ap \\ \\ \sf \to \: s_{n} = \frac{n}{2}(2a + (n - 1)d)\end{gathered}
→
somerelatedformula
→1)=ntermsofanap
→a
n
=a+(n−1)d
→2)=sumofnthtermsofanap
→s
n
=
2
n
(2a+(n−1)d)
\begin{gathered}\sf \to \: nth \: term \: of \: an \: gp \\ \\ \sf \to \: t_{n} = ar {}^{n - 1} \\ \\ \sf \to \: sum \: of \: nth \: terms \: of \: an \: gp \\ \\ \sf \to \: s_{n} = [tex[\dfrac{a( {r}^{n} - 1)}{(r - 1)}\end{gathered}
→nthtermofangp
→t n
=ar n−1
→sum of nth terms of angp
→s n = (r−1)a(r n −1)
Common difference
=> Sn = n/2 ( 2a + ( n - 1 ) d
=> 400 = 16/2 ( 2(5) + 15d )
=> 400 = 8 ( 10 + 15d )
=> 400 = 80 + 120d
=> 320 = 120d
=> d = 320 / 120 = 8/3
I hope it will be helpful for you.
