Question

If a₁, a₂, a₃,..., an are in A. P. with a₁ = 3, an, = 39 and a₁+a₂+...+an = 210, then the value of n is equal to

• A. 8
• B. 10
• C. 11
• D. 13
• E. 15

Answer 1

The Correct Option is B

We are given that \(a_1, a_2, a_3, \dots, a_n\) are in arithmetic progression (A.P.), with \(a_1 = 3 ,  a_n = 39 , \text{and }\ a_1 + a_2 + \dots + a_n = 210 \) We need to find the value of \(n \)

We know the sum of the first \(n\) terms of an A.P. is given by the formula:

\(S_n = \frac{n}{2} \left( 2a_1 + (n-1) d \right)\)

where \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the number of terms.

We are also given that the sum of the terms is 210, so:

\(\frac{n}{2} \left( 2a_1 + (n-1) d \right) = 210\)

Substitute \(a_1 = 3\) into the equation:

\(\frac{n}{2} \left( 6 + (n-1) d \right) = 210\)

\(n(6 + (n-1) d) = 420\)

We also know that \(a_n = a_1 + (n-1) d\), and \(a_n = 39 , s\)

\(39 = 3 + (n-1) d\)

\(36 = (n-1) d\)

\(d = \frac{36}{n-1}\)

Now substitute this value of \(d\) into the earlier equation:

\(n \left( 6 + (n-1) \cdot \frac{36}{n-1} \right) = 420\)

\(n(6 + 36) = 420\)

\(42n = 420\)

\(n = 10\)

The value of \( n \) is 10.