Question

what is the fourth term of an AP of n terms whose sum is n(n+1)​

Answer 1

Answer:

Forth term of the A.P is $\huge{\boxed{\bold{\mathsf{A_4=8}}}}$

Step-by-step explanation:

Given:

Arithmetic progression with "n" numbers of terms.

whose sum $\bold{\mathsf{S_n= n(n+1)}}$......(i)

To find : fourth term $\bold{\mathsf{A_4 = a + (n-1)d }}$

Let first term be a , common difference be d.

Put n = 1 in eqn (i)

$\bold{\mathsf{S_n= n(n+1)}}$

$\bold{\mathsf{S_1= 1(1+1)}}$

$\bold{\mathsf{S_1= 2}}$

put n = 2 in eqn (i)

$\bold{\mathsf{S_n= n(n+1)}}$

$\bold{\mathsf{S_2= 2(2+1)}}$

$\bold{\mathsf{S_2= 6}}$

As we know $\bold{\mathsf{S_1= }}$ that is sum of 1st term will be its the first term.

so, $\bold{\mathsf{A_1= 2}}$

Now $\bold{\mathsf{S_2=A_1 + A_2 }}$

Now put known values

$\bold{\mathsf{6=2 + A_2 }}$

$\bold{\mathsf{A_2=4 }}$

we get A2 = 4 , A1 = 2

$\bold{\mathsf{A_2 - A_1 =4 -2 = 2}}$

Apply formula

$\bold{\mathsf{A_n=a + (n-1)d}}$

Put n= 4

Put in all the known values

$\bold{\mathsf{A_4=a + (4-1)d}}$

$\bold{\mathsf{A_4=8}}$

hence, Forth term of the A.P is $\bold{\mathsf{A_4=8}}$

$\rule{200}{1}$

Formula used:-

$\bold{\mathsf{A_n=a + (n-1)d}}$