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

Question:

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

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