An AP 5, 12, 19, ... has 50 terms. Find its last term. Hence, find the sum of its last 15 terms.

Solution:

(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.

$l=a_{50}=5+(50-1) \times 7 \quad\left[a_{n}=a+(n-1) d\right]$

$=5+343$

$=348$

Thus, the last term of the AP is 348.

Now,

Sum of the last 15 terms of the AP

$=S_{50}-S_{35}$

$=\frac{50}{2}[2 \times 5+(50-1) \times 7]-\frac{35}{2}[2 \times 5+(35-1) \times 7] \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$

$=\frac{50}{2} \times(10+343)-\frac{35}{2} \times(10+238)$

$=\frac{50}{2} \times 353-\frac{35}{2} \times 248$

$=\frac{17650-8680}{2}$

$=\frac{8970}{2}$

$=4485$

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.

$l=a_{60}=8+(60-1) \times 2 \quad\left[a_{n}=a+(n-1) d\right]$

$=8+118$

$=126$

Thus, the last term of the AP is 126.

Now,

Sum of the last 10 terms of the AP

$=S_{60}-S_{50}$

$=\frac{60}{2}[2 \times 8+(60-1) \times 2]-\frac{50}{2}[2 \times 8+(50-1) \times 2] \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$

$=30 \times(16+118)-25 \times(16+98)$

$=30 \times 134-25 \times 114$

$=4020-2850$

$=1170$

Hence, the required sum is 1170.