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

Entered answer:

✅ Correct Answer: 65

For an AP with first term $a$ and common difference $d$ :

$a_n = a + (n-1)d$

$S_n = \dfrac{n}{2}[2a + (n-1)d]$

Given $a_4 + a_7 + a_{10} = 99$ :

$(a + 3d) + (a + 6d) + (a + 9d) = 99$

$3a + 18d = 99$

$a + 6d = 33 \quad \cdots (i)$

Given $S_{14} = 497$ :

$\dfrac{14}{2}[2a + 13d] = 497$

$7(2a + 13d) = 497$

$2a + 13d = 71 \quad \cdots (ii)$

From equation $(i)$ : $a = 33 - 6d$

Substituting in equation $(ii)$ :

$2(33 - 6d) + 13d = 71$

$66 - 12d + 13d = 71$

$d = 5$

$a = 33 - 6(5) = 3$

$S_5 = \dfrac{5}{2}[2(3) + 4(5)]$

$= \dfrac{5}{2}[6 + 20]$

$= \dfrac{5}{2} \times 26$

$= 65$

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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

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