Question

(i) An AP 5, 12, 19, ... has 50 terms. Find its last term. Hence, find the sum of its last 15 terms. [CBSE 2015]

(ii) An AP 8, 10, 12, ... has 60 terms. Find its last term. Hence, find the sum of its last 10 terms. [CBSE 2015]

Answer 1

(i) The given AP is 5, 12, 19, ... .

Here, a = 5, d = 12 − 5 = 7 and n = 50.

Since there are 50 terms in the AP, so the last term of the AP is a₅₀.

$$\begin{gathered} l=a_{50}=5+\left(50-1\right)\times 7\left[a_n=a+\left(n-1\right)d\right] \\ =5+343 \\ =348 \end{gathered}$$

Thus, the last term of the AP is 348.

Now,

Sum of the last 15 terms of the AP

$$\begin{gathered} =S_{50}-S_{35} \\ =\frac{50}{2}\left[2\times 5+\left(50-1\right)\times 7\right]-\frac{35}{2}\left[2\times 5+\left(35-1\right)\times 7\right]\left\{S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\right\} \\ =\frac{50}{2}\times \left(10+343\right)-\frac{35}{2}\times \left(10+238\right) \\ =\frac{50}{2}\times 353-\frac{35}{2}\times 248 \end{gathered}$$

$$\begin{gathered} =\frac{17650-8680}{2} \\ =\frac{8970}{2} \\ =4485 \end{gathered}$$

Hence, the required sum is 4485.

(ii) The given AP is 8, 10, 12, ... .

Here, a = 8, d = 10 − 8 = 2 and n = 60

Since there are 60 terms in the AP, so the last term of the AP is a₆₀.

$$\begin{gathered} l=a_{60}=8+\left(60-1\right)\times 2\left[a_n=a+\left(n-1\right)d\right] \\ =8+118 \\ =126 \end{gathered}$$

Thus, the last term of the AP is 126.

Now,

Sum of the last 10 terms of the AP

$$\begin{gathered} =S_{60}-S_{50} \\ =\frac{60}{2}\left[2\times 8+\left(60-1\right)\times 2\right]-\frac{50}{2}\left[2\times 8+\left(50-1\right)\times 2\right]\left\{S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\right\} \\ =30\times \left(16+118\right)-25\times \left(16+98\right) \\ =30\times 134-25\times 114 \end{gathered}$$
$$\begin{gathered} =4020-2850 \\ =1170 \end{gathered}$$

Hence, the required sum is 1170.