The first term of an AP is 6 and the fifth term is 18. Find the number of terms in the series having a sum of 162

Answer 1

Let the common difference be d.

Then the second term is 6+d, the third term is 6+2d, the fourth term is 6+3d, and the fifth term is 6+4d.

So we have the equation 6 + 4d = 18.

Solving for d, we find that d = 3.

Now we can write out the terms of the AP: 6, 9, 12, 15, 18, ...

The sum of the first n terms of an AP is given by the formula: S_n = (n/2)(a_1 + a_n), where a_1 is the first term and a_n is the nth term.

We want to find the number of terms, n, such that S_n = 162.

Plugging in the values for a_1 and a_n, we have: 162 = (n/2)(6 + a_n).

We can find a_n by substituting n into the nth term formula: a_n = a_1 + (n-1)d.

Plugging in the values for a_1 and d, we have: a_n = 6 + (n-1)3 = 6 + 3n - 3 = 3n + 3.

Now we can substitute this into the equation for S_n: 162 = (n/2)(6 + 3n + 3).

Simplifying, we get: 162 = (n/2)(3n + 9).

Dividing both sides by 3, we have: 54 = (n/2)(n + 3).

Expanding the equation, we get: 54 = (n^2 + 3n)/2.

Multiplying both sides by 2, we have: 108 = n^2 + 3n.

Rearranging the equation, we get: n^2 + 3n - 108 = 0.

Factoring the quadratic, we have: (n - 9)(n + 12) = 0.

So we have two possible solutions: n = 9 and n = -12.

Since the number of terms cannot be negative, the number of terms is 9.