The sum of the first five terms of an AP and | vedclass

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Question

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

Vedclass · Accepted Answer

$970$

Vedclass · Answer

(C) Let the first term be $a$ and the common difference be $d$ of an $AP$.The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.For the first five terms: $S_5 = \frac{5}{2}[2a + 4d] = 5(a + 2d) = 5a + 10d$.For the first seven terms: $S_7 = \frac{7}{2}[2a + 6d] = 7(a + 3d) = 7a + 21d$.Given $S_5 + S_7 = 167$,we have $(5a + 10d) + (7a + 21d) = 167$,which simplifies to $12a + 31d = 167$ (Eq. $1$).Given the sum of the first ten terms $S_{10} = 235$,we have $\frac{10}{2}[2a + 9d] = 235$,which simplifies to $5(2a + 9d) = 235$,or $2a + 9d = 47$ (Eq. $2$).Multiplying Eq. $2$ by $6$,we get $12a + 54d = 282$ (Eq. $3$).Subtracting Eq. $1$ from Eq. $3$: $(12a + 54d) - (12a + 31d) = 282 - 167$,which gives $23d = 115$,so $d = 5$.Substituting $d = 5$ into Eq. $2$: $2a + 9(5) = 47 \Rightarrow 2a + 45 = 47 \Rightarrow 2a = 2 \Rightarrow a = 1$.The sum of the first twenty terms is $S_{20} = \frac{20}{2}[2a + (20-1)d] = 10[2(1) + 19(5)] = 10[2 + 95] = 10[97] = 970$.

Column $A$	Column $B$
$(A_{1}) \quad 2, -2, -6, -10, \ldots$	$(B_{1}) \quad \frac{2}{3}$
$(A_{2}) \quad a = -18, n = 10, a_{n} = 0$	$(B_{2}) \quad -5$
$(A_{3}) \quad a = 0, a_{10} = 6$	$(B_{3}) \quad 4$
$(A_{4}) \quad a_{2} = 13, a_{4} = 3$	$(B_{4}) \quad -4$
	$(B_{5}) \quad 2$
	$(B_{6}) \quad \frac{1}{2}$
	$(B_{7}) \quad 5$

The sum of the first five terms of an $AP$ and the sum of the first seven terms of the same $AP$ is $167$. If the sum of the first ten terms of this $AP$ is $235$,find the sum of its first twenty terms.

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