If the sum of the first 6 terms of an AP is 3 | vedclass

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(A) Let $a$ be the first term and $d$ be the common difference of the $AP$.The sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) Let $a$ be the first term and $d$ be the common difference of the $AP$.The sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Given $S_6 = 36$:$\frac{6}{2}[2a + (6-1)d] = 36$$3[2a + 5d] = 36$$2a + 5d = 12$ ....$(i)$Given $S_{16} = 256$:$\frac{16}{2}[2a + (16-1)d] = 256$$8[2a + 15d] = 256$$2a + 15d = 32$ ....$(ii)$Subtracting equation $(i)$ from equation $(ii)$:$(2a + 15d) - (2a + 5d) = 32 - 12$$10d = 20$$d = 2$Substituting $d = 2$ into equation $(i)$:$2a + 5(2) = 12$$2a + 10 = 12$$2a = 2$$a = 1$Now,find the sum of the first $10$ terms $(S_{10})$:$S_{10} = \frac{10}{2}[2a + (10-1)d]$$S_{10} = 5[2(1) + 9(2)]$$S_{10} = 5[2 + 18]$$S_{10} = 5 	imes 20 = 100$.Thus,the sum of the first $10$ terms is $100$.

If the sum of the first $6$ terms of an $AP$ is $36$ and that of the first $16$ terms is $256,$ find the sum of the first $10$ terms.

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