Find the sum to n terms of the A.P., whose kth term is 5k + 1.
It is given that the kth term of the A.P. is 5k + 1.
kth term = ak = a + (k – 1)d
∴ a + (k – 1)d = 5k + 1
a + kd – d = 5k + 1
Comparing the coefficient of k, we obtain d = 5
a – d = 1
⇒ a – 5 = 1
⇒ a = 6
\begin{align} S_n = \frac{n}{2}\left[2a + (n-1)d\right] \end{align}
\begin{align} = \frac{n}{2}\left[2(6) + (n-1)(5)\right] \end{align}
\begin{align} = \frac{n}{2}\left[12 + 5n -5\right] \end{align}
\begin{align} = \frac{n}{2}\left(5n + 7\right) \end{align}
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2 Comment(s) on this Question
How the compare the coefficient of k.
how do you compare the coefficients in that given question???
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