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

Approach: In any AP, three terms equally spaced around a middle term sum to three times that middle term, and a sum of \(n\) terms is \(n\) times the middle (average) term. Use this to read off the AP quickly.

Step 1: Let the first term be \(a\) and common difference \(d\), so \(T_n = a + (n-1)d\). The 4th, 7th and 10th terms are symmetric about the 7th, so \[ T_4 + T_7 + T_{10} = 3\,T_7 = 99 \implies T_7 = a + 6d = 33. \]

Step 2: Sum of first 14 terms: \[ S_{14} = \frac{14}{2}\left(2a + 13d\right) = 7(2a + 13d) = 497 \implies 2a + 13d = 71. \]

Step 3: Solve the two equations. From \(a + 6d = 33\), multiply by 2: \(2a + 12d = 66\). Subtract from \(2a + 13d = 71\): \[ d = 5, \qquad a = 33 - 6(5) = 3. \]

Step 4: Sum of first 5 terms: \[ S_5 = \frac{5}{2}\left(2a + 4d\right) = \frac{5}{2}\left(6 + 20\right) = \frac{5}{2}\times 26 = 65. \]

Final answer: \(S_5 = \boxed{65}\).

Why this works: Spotting that \(T_4 + T_7 + T_{10} = 3T_7\) gives the middle term for free, avoiding a messy substitution.

Answer 2

Alternate approach — using the middle-term property of an AP:
In an AP, the 4th, 7th and 10th terms are equally spaced around the 7th term, so their sum equals 3 times the 7th term: \( 3(a+6d)=99 \Rightarrow a+6d=33 \).
Also, the sum of the first 14 terms equals 7 times the sum of the 7th and 8th terms (pairing terms symmetrically from both ends): \( 7(a+6d + a+7d) = 497 \Rightarrow 2a+13d=71 \).
From \( a+6d=33 \): \( a=33-6d \). Substituting: \( 2(33-6d)+13d=71 \Rightarrow d=5, a=3 \).
Sum of first 5 terms \( =\frac{5}{2}(2a+4d)=\frac{5}{2}(6+20)=\) 65.