# Find the sum of the first 15 terms of each of the following sequences having nth term as

Question:

Find the sum of the first 15 terms of each of the following sequences having nth term as

(i) $a_{n}=3+4 n$

(ii) $b_{n}=5+2 n$

(iii) $x_{n}=6-n$

(iv) $y_{n}=9-5 n$

Solution:

(i) Here, we are given an A.P. whose $n^{\text {th }}$ term is given by the following expression, $a_{n}=3+4 n$. We need to find the sum of first 15 terms.

So, here we can find the sum of the $n$ terms of the given A.P., using the formula, $S_{n}=\left(\frac{n}{2}\right)(a+l)$

Where, a = the first term

l = the last term

So, for the given A.P,

The first term (a) will be calculated using in the given equation for nth term of A.P.

$a=3+4(1)$

$=3+4$

$=7$

Now, the last term (l) or the nth term is given

$l=a_{n}=3+4 n$

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

$S_{15}=\left(\frac{15}{2}\right)[(7)+3+4(15)]$

$=\left(\frac{15}{2}\right)[10+60]$

$=\left(\frac{15}{2}\right)(70)$

$=(15)(35)$

$=525$

Therefore, the sum of the 15 terms of the given A.P. is $S_{15}=525$.

(ii) Here, we are given an A.P. whose $n^{\text {th }}$ term is given by the following expression

We need $b_{n}=5+2 n$ to find the sum of first 15 terms.

So, here we can find the sum of the $n$ terms of the given A.P., using the formula,

$S_{n}=\left(\frac{n}{2}\right)(a+l)$

Where, a = the first term

l = the last term

So, for the given A.P,

The first term (a) will be calculated using in the given equation for nth term of A.P.

$b=5+2(1)$

$=5+2$

$=7$

Now, the last term (l) or the nth term is given

$l=b_{n}=5+2 n$

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

$S_{15}=\left(\frac{15}{2}\right)[(7)+5+2(15)]$

$=\left(\frac{15}{2}\right)[12+30]$

$=\left(\frac{15}{2}\right)(42)$

$=(15)(21)$

$=315$

Therefore, the sum of the 15 terms of the given A.P. is $S_{15}=315$.

(iii) Here, we are given an A.P. whose nth term is given by the following expression, . We need to find the sum of first 15 terms.

So, here we can find the sum of the n terms of the given A.P., using the formula,

$S_{n}=\left(\frac{n}{2}\right)(a+l)$

Where, a = the first term

l = the last term

So, for the given A.P,

The first term (a) will be calculated using in the given equation for nth term of A.P.

$x=6-1$

$=5$

Now, the last term (l) or the nth term is given

$l=a_{n}=6-n$

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

$S_{15}=\left(\frac{15}{2}\right)[(5)+6-15]$

$=\left(\frac{15}{2}\right)[11-15]$

$=\left(\frac{15}{2}\right)(-4)$

$=(15)(-2)$

$=-30$

Therefore, the sum of the 15 terms of the given A.P. is $S_{15}=-30$.

(iv) Here, we are given an A.P. whose $n^{\text {th }}$ term is given by the following expression, $y_{n}=9-5 n$. We need to find the sum of first 15 terms.

So, here we can find the sum of the n terms of the given A.P., using the formula,

$S_{n}=\left(\frac{n}{2}\right)(a+l)$

Where, a = the first term

l = the last term

So, for the given A.P,

The first term (a) will be calculated using in the given equation for nth term of A.P.

$y=9-5(1)$

$=9-5$

$=4$

Now, the last term (l) or the nth term is given

$l=a_{n}=9-5 n$

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

$S_{15}=\left(\frac{15}{2}\right)[(4)+9-5(15)]$

$=\left(\frac{15}{2}\right)[13-75]$

$=\left(\frac{15}{2}\right)(-62)$

$=(15)(-31)$

$=-465$

Therefore, the sum of the 15 terms of the given A.P. is $S_{15}=-465$.