Question

The first term of an Ap is 5, the last term 45 and sum
is 400. find the number of teams and the common diffence​

Answer 1

$\large \sf \underline{ \underline{ \:Given : \: \: \: }}$

First term , a = 5

Nth term or last term , An or l = 45

Sum of first n terms , S = 400

$\large \sf \underline{ \underline{ \: To \: find : \: \: \: }}$

The number of terms and the common diffence

$\large \sf \underline{ \underline{ \: Solution : \: \: \: }}$

We know that The sum of the first n terms of an AP is given by

$\sf \huge{ \pink{\fbox{ \fbox{S = \frac{n}{2} (a + l)}}}}$

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

Now , the nth term of an AP is given by

$\pink{ \sf \huge{ \fbox{ \fbox{ a_{n} = a + (n - 1)d}}}}$

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

Hence , the number of terms and the common diffence of given AP are 16 and 8/3

Answer 2

Answer:

16, 8/3

Step-by-step explanation:

Given,  a = 5 ,   l = 45 ,    S n = 400

We know that,

S n =  n 2 [ a + l ]

⇒   400 = n 2 [ 5 + 45 ]

⇒   n = 400 /25

⇒     n = 16

and

l = a + ( n − 1 ) d

⇒ 45 = 5 + ( 16 − 1 ) d

⇒ d = 40 15

⇒ d = 8 /3