Question

For an A.P. given below find t₂₀ and S₁₀.
$\frac{1}{6},\frac{1}{4},\frac{1}{3},....$

Answer 1

In the given A.P., we have:
a = $\frac{1}{6}$ and d = t₂ – t₁ = $\frac{1}{4}-\frac{1}{6}=\frac{3-2}{12}=\frac{1}{12}$
We know:
n^th term of an A.P., t_n = a + (n – 1)d
To find the 20th term of the given A.P, we need to put n = 20 in the above expression.
Thus, we have:
$$\begin{gathered} t_{20}=\frac{1}{6}+\left(20-1\right)\left(\frac{1}{12}\right) \\ =\frac{1}{6}+\frac{19}{12} \\ =\frac{2+19}{12} \\ =\frac{21}{12} \\ =\frac{7}{4} \end{gathered}$$
Now, we know:
Sum of n terms of an A.P., $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$
To find the sum of the first 10 terms, we need to put a = $\frac{1}{6}$, d = $\frac{1}{12}$ and n = 10 in the above expression.
Thus, we have:
$$\begin{gathered} S_{10}=\frac{10}{2}\left[2\times \frac{1}{6}+\left(10-1\right)\left(\frac{1}{12}\right)\right] \\ =5\left[\frac{1}{3}+\frac{9}{12}\right] \\ =5\left[\frac{4+9}{12}\right] \\ =5\times \frac{13}{12} \\ =\frac{65}{12} \end{gathered}$$