Let Sₙ denote sum of first n terms of an arithmetic progression. If mathrm{S}

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8. Sequences and Series

hard

Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $\mathrm{S}_{10}=390$ and the ratio of the tenth and the fifth terms is $15: 7$, then $S_{15}-S_5$ is equal to:

$800$

$890$

$790$

$690$

(JEE MAIN-2024)

Solution

$ \mathrm{S}_{10}=390 $

$ \frac{10}{2}[2 \mathrm{a}+(10-1) \mathrm{d}]=390 $

$ \Rightarrow 2 \mathrm{a}+9 \mathrm{~d}=78 $ $……(1)$

$ \frac{\mathrm{t}_{10}}{\mathrm{t}_5}=\frac{15}{7} \Rightarrow \frac{\mathrm{a}+9 \mathrm{~d}}{\mathrm{a}+4 \mathrm{~d}}=\frac{15}{7} \Rightarrow 8 \mathrm{a}=3 \mathrm{~d} $ $……(2)$

$ \text { From }(1) \&(2) \quad \mathrm{a}=3 \& \mathrm{~d}=8 $

$ \mathrm{~S}_{15}-\mathrm{S}_5=\frac{15}{2}(6+14 \times 8)-\frac{5}{2}(6+4 \times 8) $

$ =\frac{15 \times 118-5 \times 38}{2}=790$

Standard 11

Mathematics

If $a_1 , a_2, a_3, . . . . , a_n, ….$ are in $A.P.$ such that $a_4 – a_7 + a_{10}\, = m$, then the sum of first $13$ terms of this $A.P.$, is ………….. $\mathrm{m}$

hard

(JEE MAIN-2013)

After inserting $n$, $A.M.'s$ between $2$ and $38$, the sum of the resulting progression is $200$. The value of $n$ is

easy

$150$ workers were engaged to finish a piece of work in a certain number of days. $4$ workers dropped the second day, $4$ more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in which the work was completed is

medium

If ${S_1},\;{S_2},\;{S_3},………..{S_m}$ are the sums of $n$ terms of $m$ $A.P.'s$ whose first terms are $1,\;2,\;3,\;……………,m$ and common differences are $1,\;3,\;5,\;………..2m – 1$ respectively, then ${S_1} + {S_2} + {S_3} + …….{S_m} = $

medium

Let $S_n$ be the sum to n-terms of an arithmetic progression $3,7,11, \ldots \ldots$. . If $40<\left(\frac{6}{\mathrm{n}(\mathrm{n}+1)} \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}\right)<42$, then $\mathrm{n}$ equals

hard

(JEE MAIN-2024)

Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $\mathrm{S}_{10}=390$ and the ratio of the tenth and the fifth terms is $15: 7$, then $S_{15}-S_5$ is equal to:

Solution

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