Question

The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 400. Find the number of terms and the common difference of the A.P.

Answer 1

Answer:

Values of n = 16 and

Values of n = 16 and d = 8/3

Step-by-step explanation:

Let a , d , l are first term , common difference and last term of an A.P , respectively.

given ,

a = 5, l = 45

We know that,

$\boxed {last \:term \:(l)=a+(n-1)d }$

Substitute values of a and l , we get,

=> 45 = 5+(n-1)d

=> 45-5 = (n-1)d

=> (n-1)d=40 ----(1)

$\boxed {Sum\: of\: n\: terms \:(S_{n})\\= \frac{n}{2}(a+l)}$

$S_{n}=400$\* given *\

$\implies \frac{n}{2}(5+45)=400$

$\implies n\times \frac{50}{2}=400$

$\implies n = 400 \times \frac{2}{50}$

$\implies n = 8 \times 2$

$\implies n = 16$

Now ,

Substitute n=16 in equation (1), we get

(16-1)d = 40

=> 15d =40

=> d = 40/15

=> d = 8/3

Therefore,

Values of n = 16 and

d = 8/3

••••

Answer 2

Answer:

The number of terms in the series is 16 and the common difference is 8/3.

Given: a = 5, l=45 and $\mathrm{S}_{\mathrm{n}}$ = 400

To find: n=? d=?

Solution: l=45

a+(n-1)d=45 (since,  $a_{n}$ = l = a+(n-1)d

l=45 …..(i)

Also,

$\begin{array}{l}{\mathrm{S}_{\mathrm{n}}=\frac{n}{2}[2 a+(n-1) d]=400} \\ {\mathrm{S}_{\mathrm{n}}=\frac{n}{2}[a+l]=400 \ldots \ldots \text { (ii) }}\end{array}$

Substituting the values in these equations and after solving the two equations, we get,  

n = 16

Also, since last term is given by =  $a_{n}$ =a+(n-1)d= 45

$\begin{array}{l}{\Rightarrow 5+(16-1) d=45} \\ {5+15 d=45} \\ {15 d=40} \\ {\Rightarrow d=\frac{40}{15}} \\ {d=\frac{8}{3}}\end{array}$

Therefore, n=16 and d = 8/3