Question

In an ap 1+4+7+10+.............+X=287,then find the value of X

Answer 1

Step-by-step explanation:

a= 1, d= 3, Sn = 287, n=?, Tn=X

Sn= n/2[2a+(n-1) d]

287 = n/2[2*1+(n-1)3]

287*2=n(2+3n-1)

574=n+3n^2

or

3n^2+n-574=0

Factorise above equation by completing square method.

You will get n=14

Tn= a + (n-1) d

X = 1 + (14-1)3

X= 1+39

X = 40

So the value of X is 40.

Answer 2

The value of X is 40.

Step-by-step explanation :

Given that, 1 + 4 + 7 + 10 + ... + X = 287

If we consider the given AP (1 + 4 + ... + X) having n terms, then X is the nth term of the AP and the sum of the n terms is 287.

Now, the sum of n terms = 287

i.e., n/2 * [2 * 1 + (n - 1) * 3] = 287

or, n/2 * [2 + 3n - 3] = 287

or, n/2 * [3n - 1] = 287

or, n (3n - 1) = 574

or, 3n² - n - 574 = 0

or, (n - 14) (3n + 41) = 0

Either n - 14 = 0 or, 3n + 41 = 0

Since n is a positive integer, n = 14

Again, we have nth term = X

i.e., 1 + (14 - 1) * 3 = X

or, X = 1 + 13 * 3

or, X = 1 + 39

or, X = 40

Therefore the value of X is 40.

Related problem :

If tn represents nth term of an AP, t2 + t5 - t3 = 10 and t2 + t9 = 17, find its first term and its common difference. - https://brainly.in/question/15685301