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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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Natural numbers are divided into groups $\;(1)\;,\;(2,3,4)\;,\;(5,6,7,8,9)\;.....$ Sum of first and last term of $n^{th}\;$ group will be :

$(a)\;n^2+n\qquad(b)\;n^2-n+2\qquad(c)\;2n^2-2n+2\qquad(d)\;n^2-2n+2$

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  • Sum of $n$ terms of an $A.P.$ with first term $a$ and common difference, $d$ is $\large\frac{n}{2}$$(2a+(n-1)d)$
Answer : (c) $\;2n^2-2n+2$
1. The numbers in the $n^{th}$ group are in $A.P.$ with common difference 1.
2. There are $2n-1$ numbers in $n^{th}$ group.
Step 1:
To find the first term in $n^{th}$ group:
The first term in each group are the sequence $ 1,2,5,10..........t_n$
$\therefore$ The first term of $n^{th}$ group is $n^{th}$ term $(t_n)$ of this sequence.
Let $S=1+2+5+10+................t_n+0$.........(i)
$\:\:\:\:\:S=\:\:0+1+2+5+10+..............+t_n$......(ii)
Subtracting (i)-(ii) we get
$0=1+(1+3+5+...........(n-1)terms)-t_n$
$\Rightarrow\:t_n=1+(1+3+5+.......(n-1)terms)$
$1+3+5+..........(n-1)terms$ is sum of $n-1$ terms of $A.P.$ with $d=2$
$=\large\frac{n-1}{2}$$(2\times1+(n-2)\times 2)=(n-1)(n-1)=(n-1)^2$
$\Rightarrow\:t_n=1+\large\frac{n-1}{2}$$(2+(n-2)2)$
$\Rightarrow\:t_n=1+(n-1)^2=n^2-2n+2$
$i.e., $ The first term of $n^{th}$ group is $n^2-2n+2$
Step 2:
To find the last term of the $n^{th}$ group:
We know that in $n^{th}$ group there are $2n-1$ terms and the numbers are in $A.P.$ with $d=1$
and with first term $t_n$
$\therefore $ The last term in the $n{th}$ group is $(2n-1)^{th}$ term in the series of $n^{th}$ group,
$i.e.,\:\:tn,(t_n+1),(t_n+2)...........$
$(2n-1)^{th}$ term of an $A.P.$ is $a+(2n-2)d$
which is $t_n+(2n-2)\times 1=n^2-2n+2+2n-2=n^2$
$\therefore$ The sum of first and last term of $n^{th}$ group is $ n^2-2n+2+n^2=2n^2-2n+2$
answered Jan 21, 2014 by yamini.v
edited Jan 24, 2014 by pady_1
 

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