Question

Solve the equation :1 + 5 + 9 + 13 + ... + x = 1326​

Answer 1

Answer:

x = 101

Step-by-step explanation:

series is in AP

Sum of AP is

S = N/2(2a + (N-1)d)

1326= N/2(2 + 4N - 4)

2652= 4N^2 - 2N

it becomes quadratic in N

4N^2 - 2N -2652= 0

4N^2 - 104N + 102N -2652=0

4N( N- 26) +102(N-26) =0

from the above we will get N= 26

last term will be (a +25d)= 101

Answer 2

The value of x is 101.

It is 26th term of the series.

Given:

• A series :1 + 5 + 9 + 13 + ... + x
• Sum of series is 1326.

To find:

• Find the value of x.

Solution:

Formula to be used:

• General term of A.P.: $\bf a_n = a + (n - 1)d$
• Sum of first n terms of A.P.:$S_n = \frac{n}{2} \left (2a + (n - 1)d \right )$ here, a: first term, d: common difference, n: number of terms

Step 1:

To find value of x, first find number of terms.

Put the values in formula of sum.

As,

a=1

d=5-1=4

Sn=1326

$1326 = \frac{n}{2} (2 \times 1 + (n - 1)4)$

or

$1326 = \frac{\cancel2n}{ \cancel2} ( 1+2 (n - 1))$

or

$1326 = n( 1+2n - 2))$

or

$\bf 2 {n}^{2} - n - 1326 = 0$

Step 2:

Solve the quadratic equation in n, to find the values of n.

Split the middle term to factorise.

$2 {n}^{2} - 52n + 51n - 1326 = 0$

or

$2n(n - 26) + 51(n - 26) = 0$

or

$(n - 26)(2n + 51) = 0$

or

$\bf n = 26$

or

$n = \frac{ - 51}{2}$

(it can't be a term of AP, as it is rational and negative)

Thus,

Total terms of A.P. are 26.

Step 3:

Find the 26 th term of A.P. , which is value of x.

$a_{26} = 1 + (26 - 1)4$

or

$a_{26} = 1 + 25 \times 4$

or

$a_{26}= 101$

Thus,

Value of x = 101.

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