Question

Find the sum of 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.

Hint:

For A.P. problems, always identify the first term and common difference from the given terms before applying the sum formula.

Answer 1

Step 1: Let the first term and common difference be \(a\) and \(d\). 
The general term of an A.P. is \(a_n = a + (n - 1)d\). 
Step 2: Use the given conditions. 
\[ a_2 = a + d = 14 \quad \text{(i)} \] \[ a_3 = a + 2d = 18 \quad \text{(ii)} \] Step 3: Subtract (i) from (ii). 
\[ (a + 2d) - (a + d) = 18 - 14 \Rightarrow d = 4 \] Step 4: Substitute \(d = 4\) in (i). 
\[ a + 4 = 14 \Rightarrow a = 10 \] Step 5: Use the sum formula of an A.P. 
\[ S_n = \dfrac{n}{2}[2a + (n - 1)d] \] For \(n = 51\): \[ S_{51} = \dfrac{51}{2}[2(10) + 50(4)] \] \[ = \dfrac{51}{2}[20 + 200] = \dfrac{51}{2} \times 220 = 51 \times 110 = 5610 \] Step 6: Conclusion. 
Hence, the sum of 51 terms of the A.P. is 5610.