Sequences and Series
If $$a_1, a_2, a_3,....a_n$$ are in A.P. and $$a_1+ a_4 + a_7 + ..... + a_{16} = 114$$, then $$a_1 + a_6 + a_{11} + a_{16}$$ is equal to:
Let $$a_{1},\ a_{2},\ a_{3},\ \ldots,\ a_{100}$$ be an arithmetic progression with $$a_{1}=3$$ and $$S_{p}$$ is sum of 100 terms . For any integer $$n$$ with $$1\leq n \leq 20$$, let $$ m=5n$$. If $$\dfrac{S_{m}}{S_{n}}$$ does not depend on $$n$$, then $$a_{2}$$ is
We know that sum of n terms of an A.P. is given by
$${ S }_{k }=\dfrac { k }{ 2 } \left\{ 2a+(k-1)d \right\} $$
So,$$\displaystyle \dfrac { { S }_{ m } }{ { S }_{ n } } =\frac { { S }_{ 5n } }{ { S }_{ n } }=\dfrac { \dfrac { 5n }{ 2 } \left\{ 6+(5n-1)d \right\} }{ \dfrac { n }{ 2 } \left\{ 6+(n-1)d \right\} } =\frac { 5n\left\{ 6+(5n-1)d \right\} }{ n\left\{ 6+(n-1)d \right\} } =\frac { 5\left\{ (6-d)+5nd \right\} }{ \left\{ (6-d)+n \right\} } $$
Since, $$\displaystyle \frac { { S }_{ m } }{ { S }_{ n } }$$ is independent of n , so $$ d =6$$.
If $$d = 6$$ then $${a}_{2} = {a}_{1} + d = 3 + 6 = 9$$
If $$a_1, a_2, a_3,....a_n$$ are in A.P. and $$a_1+ a_4 + a_7 + ..... + a_{16} = 114$$, then $$a_1 + a_6 + a_{11} + a_{16}$$ is equal to:
If the sum of the first n terms of the series $$\sqrt{3}+\sqrt{75}+\sqrt{243}+\sqrt{507}+....$$ is $$435\sqrt{3}$$, then n equals.
$$\displaystyle{ a }_{ 1 }=3,{ a }_{ n }={ 3a }_{ n-1 }+2$$ for all $$ n>1$$
Find the sum of odd integers from $$1$$ to $$2001$$.
Find the sum to $$n$$ terms of the $$A.P.$$, whose $$k^{th}$$ term is $$5k+1$$.
If the sum of first $$p$$ terms of an A.P.is equal to the sum of the first $$q$$ terms, then find the sum of the first $$(p+q)$$ terms.
Sum of the first $$p, q$$ and $$r$$ terms of an A.P. are $$a,b$$ and $$c$$ respectively. Prove that $$\displaystyle \frac { a }{ p } (q-r)+\frac { b }{ q } (r-p)+\frac { c }{ r } (p-q)=0$$
Between $$1$$ and $$31$$, $$m$$ numbers have been inserted in such a way that the resulting sequence is an $$A.P$$. and the ratio of $$\displaystyle { 7 }^{ th }$$ and $$(n-1)^{th}$$ numbers is $$5 : 9$$ . Find the value of $$n$$.
$$\displaystyle 1\times 2+2\times 3+3\times 4+4\times 5+...$$
$$\displaystyle 1\times 2\times 3 +2\times 3\times 4+3 \times 4\times 5+...$$