in an A.P. 5,9,13,17,..., FIND THE VALUE OF N FOR WHICH THE SUM OF FIRST 2n terms will exceed the sum of the first n terms by 234.

The sum of an arithmetic progression can be calculated using the formula:

S_n = n/2 * (2a + (n-1)d)

where: S_n is the sum of the first n terms, a is the first term, d is the common difference, and n is the number of terms.

Given the arithmetic progression 5, 9, 13, 17, ..., we can see that a = 5 and d = 4.

The problem states that the sum of the first 2n terms exceeds the sum of the first n terms by 234. We can express this as:

S_2n - S_n = 234

Substituting the formula for S_n into this equation gives:

2n/2 * (2*5 + (2n-1)4) - n/2 * (25 + (n-1)*4) = 234 n * (20 + 8n - 4) - n * (10 + 4n - 4) = 234 n * (16n + 16) - n * (4n + 6) = 234 16n^2 + 16n - 4n^2 - 6n = 234 12n^2 + 10n - 234 = 0

This is a quadratic equation in the form ax^2 + bx + c = 0, where a = 12, b = 10, and c = -234. We can solve for n using the quadratic formula:

n = [-b ± sqrt(b^2 - 4ac)] / (2a) n = [-10 ± sqrt((10)^2 - 412(-234))] / (2*12) n = [-10 ± sqrt(100 + 11232)] / 24 n = [-10 ± sqrt(11332)] / 24 n = [-10 ± 106.45] / 24

This gives two possible solutions: n = 4.02 or n = -4.02. However, n must be a positive integer in the context of this problem, so we round 4.02 up to the nearest whole number.

Therefore, the value of n for which the sum of the first 2n terms will exceed the sum of the first n terms by 234 is n = 5.