Question

In an arithmetic progression, if the sum of fourth, seventh and tenth terms is 99, and the sum of the first fourteen terms is 497, then the sum of first five terms is

Hint:

When given conditions on specific terms and on the sum of terms in an AP, convert them into equations using $T_n = a + (n-1)d$ and $S_n = \frac{n}{2}[2a + (n-1)d]$. Two independent conditions will usually give you two linear equations in $a$ and $d$.

Answer 1

Correct Answer: 65

Step 1: Define variables.  
Let the first term of the AP be a and the common difference be d. The n-th term:

T_n = a + (n-1)d

The sum of the first n terms:

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

Step 2: Use the condition on the 4th, 7th and 10th terms. 
Given:

T₄ + T₇ + T₁₀ = 99.

Now:

T₄ = a + 3d, T₇ = a + 6d, T₁₀ = a + 9d.

So:

(a + 3d) + (a + 6d) + (a + 9d) = 99 
3a + 18d = 99 
a + 6d = 33. (1)

Step 3: Use the condition on the sum of first 14 terms. 
Given:

S₁₄ = 497.

Using the sum formula:

S₁₄ = (14/2) [2a + (14-1)d] 
497 = 7 [2a + 13d] 
2a + 13d = 71. (2)

Step 4: Solve the system of equations for a and d. 
From (1):

a = 33 - 6d.

Substitute into (2):

2(33 - 6d) + 13d = 71 
66 - 12d + 13d = 71 
66 + d = 71 
d = 71 - 66 = 5.

Now back-substitute into (1):

a + 6(5) = 33 
a + 30 = 33 
a = 3.

So the AP begins:

3, 8, 13, 18, ...

Step 5: Find the sum of the first five terms. 
Use the sum formula with n = 5:

S₅ = (5/2) [2a + (5-1)d] 
= (5/2) [2(3) + 4(5)] 
= (5/2) [6 + 20] 
= (5/2) * 26 
= 5 * 13 = 65.

Thus, the sum of the first five terms is 65.