Question

sum of first n terms is sn =4n2+n find PA​

Answer 1

Question:

Sum of first n terms of an AP is $S_{n}= 4{n}^{2}+n$ . Find AP

Answer:

The AP is : 5 , 13 , 21 , 29 , .........

Given:

Sum of first n terms is is $S_{n}= 4{n}^{2}+n$.

To Find:

The AP

Solution:

We are given,

Sum of first n terms is is $S_{n}= 4{n}^{2}+n$.

$: \implies \rm S_{1} = 4{1}^{2} + 1$

$: \implies \rm S_{1} = 4{1}^{2} + 1$

$: \implies \rm S_{1} = 4 + 1$

$: \implies \rm S_{1} = 5 \longrightarrow a \longrightarrow t_{1}$

$: \implies \rm S_{2} = 4{2}^{2} + 2$

$: \implies \rm S_{2} = ( 4 \times 4) + 2$

$: \implies \rm S_{2} = 16 + 2$

$: \implies \rm S_{2} = 18$

but,

${\implies{\blue{\boxed{\red{\star{\rm t_{1}+ t_{2} = sum\:of\:first\:two\:terms}}}}}}$

$\therefore \rm t_{1}+ t_{2} = 18$

$\longrightarrow \rm 5 + t_{2} = 18$

$\longrightarrow \rm t_{2} = 18 - 5$

$\longrightarrow t_{2} = 13$

we know,

${\implies{\blue{\boxed{\red{\star{\rm t_{2}+ t_{1} = d}}}}}}$

where,

t(1) = first term of AP

t(2) = second term of AP

d = difference between two terms of AP.

$\therefore \longrightarrow 13 - 5 = d$

${\longrightarrow{ \red{ \boxed{ \blue{\rm d = 8 }}}}}$

Thus the required AP =

a, ( a+d ) , ( a + 2d ),..............

Thus the required AP = 5 , ( 5 + 8 ) , ( 5 + { 2 × 8})........

The required AP = 5 ,13 ,21 ,29, .......