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:

Step-by-step explanation:

Given :-

a = 5, a(n) = 45, S(n)= 400

To Find :-

Number of Terms

Common Difference

Formula to be used :-

$\boxed {(S_{n})=\frac{n}{2}\times[a+a_{n}]}$

$\boxed {a_{n}=a+(n-1)\times d}$

Solution :-

$(S_{n})=\frac{n}{2}\times[a+a_{n}]$

$\implies400=\frac{n}{2}\times[5+45]$

$\implies400=\frac{n}{2}\times[50]$

$\implies25n=400$

$\implies n=\frac{400}{25}$

$\implies n = 16$

Now, $a_{n}=a+(n-1)\times d$

$\implies 45=5+(16-1)\times d$

$\implies 45 - 5=15d$

$\implies 15d=40$

$\implies d =\frac{8}{3}$

Hence, Number of Terms are 16 and the common difference is $\frac{8}{3}.$

Answer 2

$\bold{ \large{ \underline{ \underline{ \: Answer : \: \: \: }}}}$

$\to$ Number of terms ( n ) = 16

$\to$ Common difference ( d ) = 8/3

$\bold{ \large{ \underline{ \underline{ \:Explaination : \: \: \: }}}}$

Given ,

First term ( a ) = 5

Last term ( l ) = 45

Sum of nth term ( Sn ) = 400

We know that ,

$\large{ \fbox{ \fbox{ \bold{Sn = \frac{n}{2} (a + l)}}}}$

$\to 400= \frac{n}{2} (5 + 45) \\ \\ \to 800 = 50n \\ \\ \to n = \frac{800}{50} \\ \\ \to n = 16$

Now ,

$\large \bold{ \bold{ \fbox{ \fbox{ \: nth \: term = a + ( n - 1 )d \: \: }}}}$

$\to 45 = 5 + (16 - 1)d \\ \\ \to 45 - 5 = 15d \\ \\ \to d = \frac{40}{15} \\ \\ \to d = \frac{8}{3}$

Hence , the number of terms and common difference of the given AP is 16 and 8/3