Question

For any series 5+7+9+...... the sum of how many terms will be 480 ?

Answer 1

QUESTION :-

For a series 5+7+9+.... The sum of how many terms will be 480

SOLUTION :-

Series = 5 , 7 , 9 .....

Here a1 = 5 a2 = 7 and a3 = 9

d = a2 - a1

= 7 - 5

= 2

d = a3 - a2

= 9 - 7

= 2

As the common difference is same the series is in A.P

a = 5

d = 2

sn = 480

sn = n/2 {2a + (n - 1)d}

480 = n/2 {2(5) + (n - 1)2}

480 x 2 = n {10 + 2n - 2}

960 = n {8 + 2n}

960 = 8n + 2n^2

2n^2 + 8n - 960 = 0

BY USING QUADRATIC FORMULA or FACTORIZATION METHOD

we get n = 20

Therefore 20 terms are required to get the sum of 480

Answer 2

Answer:

n = 20

Step-by-step explanation:

Given :

To find the the nth term, by which the sum till nth term of a series is 480, and the series is in A.P,

Given series(A.P) :

5, 7 , 9 , ...

Solution :

We know that the first term is 5,

All the terms have common difference 2, i.e., 7 - 5 = 9 - 7 =...

Then, we use the formula, ( for sum of first n terms of an A.P )

$S_n = \frac{n}{2}(2a + (n-1)d)$,

where,

$S_n$ indicates the sum of n terms, (Subscript n indicates the number of terms)

$a$ indicates first of the series,

$d$ indicates the common difference,

-

By subsituting the given values,

We get,

$480 = \frac{n}{2}(2(5) + (n-1)2)$

$480 = \frac{n}{2} (10 + 2n-2)$

$480 = \frac{n}{2} (2n + 8)$

$480 = \frac{n}{2} (2(n+4))$

$480 = n(n+4)$

$480 = n^2 + 4n$

$n^2 + 4n - 480 = 0$

$n^2 + 24n - 20n - 480 = 0$

$n(n-20) + 24(n-20) = 0$

$(n+24)(n+-20) = 0$

For the product to be zero,

Either,

n + 24 = 0 (or) n - 20 = 0,

n = 20 (or) n = -24, n ≠ -24, as counting numbers can't be negative (n∈N),

Hence,

n = 20,.

The sum of first 20 terms of the given series equals 480