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⚡ AZUCATION SHORTCUT:
For any AP, $a_4 + a_7 + a_{10} = 3 \times a_7$.
So, $3a_7 = 99 \implies a_7 = 33$.
Sum of 14 terms $S_{14} = \frac{14}{2}(a_1 + a_{14}) = 7(a_7 + a_8) = 497$.
$a_7 + a_8 = 71$. Since $a_7 = 33$, $a_8 = 38$. Thus, $d = 5$.

Step 1: Use the property of middle terms
Given: $a_4 + a_7 + a_{10} = 99$.
In an AP, $a_4 + a_{10} = 2a_7$.
$2a_7 + a_7 = 99 \implies 3a_7 = 99 \implies \mathbf{a_7 = 33}$.

Step 2: Solve for common difference ($d$)
Sum of first 14 terms $S_{14} = \frac{14}{2}[2a + 13d] = 497$.
$7[2a + 13d] = 497 \implies 2a + 13d = 71$.
We know $a_7 = a + 6d = 33 \implies 2a + 12d = 66$.
Subtracting the equations: $(2a + 13d) - (2a + 12d) = 71 - 66 \implies \mathbf{d = 5}$.

Step 3: Find the first term ($a$)
$a + 6(5) = 33 \implies a + 30 = 33 \implies \mathbf{a = 3}$.

Step 4: Calculate Sum of first 5 terms ($S_5$)
$S_5 = \frac{5}{2}[2(3) + 4(5)] = \frac{5}{2}[6 + 20] = \frac{5}{2} \times 26$.
$S_5 = 5 \times 13 = \mathbf{65}$.

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