# if sum of first n terms of an A.P.is 5(n)^2+3(n) find the A.P.& n^th term.?

### 5 Answers

- PuggyLv 71 decade agoFavorite Answer
So it is given that

S(n) = 5n^2 + 3n

And you want to find the arithmetic progression and the nth term.

Recall that the sum of an arithmetic progression is given by the formula:

S(n) = n(2[a1] + (n - 1)d) / 2

{Note: a1 represents the unknown first term, and d represents the difference between consecutive terms}

Through algebraic manipulation with which I won't show the details, we can replace S(n) to be

S(n) = (d/2)n^2 + n(a1 - d/2)

BUT

S(n) = 5n^2 + 3n, so all we have to do is equate the coefficients of n^2 and the coefficients of n.

d/2 = 5 (implies d = 10)

a1 - d/2 = 3 (implies a1 = 8)

Therefore, a1 (our first term) is equal to 8, and d = 10.

Our sequence then goes as follows;

8, 18, 28, 38, 48, ...

with our general term being 8 + (n - 1)10, or (8 + 10n - 10), or

(10n - 2)

Conclusion:

The arithmetic progression is the sequence with a1 = 8 and r = 10, and

a[n] = (10n - 2)

Edit:

---

If you're interested in knowing the details of how I got from

S(n) = n(2[a1] + (n - 1)d) / 2

to

S(n) = (d/2)n^2 + n(a1 - d/2)

Here it is, below.

S(n) = n(2[a1] + (n - 1)d) / 2

Factor out a 1/2, since it's a fraction anyway.

S(n) = (n/2) (2[a1] + (n - 1)d)

Expand the inside.

S(n) = (n/2) (2[a1] + nd - d)

Distribute the (n/2)

S(n) = (a1)n + d(n^2)/2 - dn/2

Group together the n^2 terms and the n terms.

S(n) = d(n^2)/2 + (a1)n - dn/2

S(n) = (d/2)n^2 + (a1 - d/2)n

It was only through this algebraic manipulation that we could equate this component-wise with our given sum.

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- 1 decade ago
Sn = 5 n^2 + 3n

S1 = 5 + 3

= 8 is the first term(a)

S2 = 5*4 + 3*2

= 26 is the sum of first two terms

second term = 26 -8

= 18

S3 = 5*9 + 9

= 54 is the sum of first three terms

third term = 54 - 26

= 28

A.P is 8,18,28,..

a= 8, d = 10

nth term = a + (n-1)d

= 8 + (n-1)10

= 10n -2

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- Mein Hoon NaLv 71 decade ago
becuase it is ap let kth term be ak+b

sum of n terms a n(n+1)/2 + bn = an^2/2 + (b+a/2) n

comparing with 5n^2 + 3n

a/2 = 5 oe a = 10 and b+5 = 3 or b = -2

nth term = 10n - 2

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- Wal CLv 61 decade ago
S_n = 5n² + 3n

S_(n - 1) = 5(n - 1)² + 3(n - 1)

= 5n² - 10n + 5 + 3n - 3

= 5n² - 7n + 2

Now S_n = S_(n - 1) + U_n

So U_n = S_n - S_(n - 1)

= 10n - 2

So the AP is {8, 18, 28, ...., 10n - 2, ....}

Source(s): Me ;^))- Login to reply the answers

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- HelmutLv 71 decade ago
Xn = 5(n)^2 + 3(n) - 5(n-1)^2 - 3n + 3

Xn = 5(n)^2 + 3(n) - 5n^2 + 10n - 5 - 3n + 3

Xn = 10n - 2

N = ∑ 8 + 18 + 28 + . . . . . + (10n - 2)

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