QUESTION BANK
1) In AP , if the common difference $(d)$ = -4 and the seventh term $a_7$ is 4, find the first term. Solution $n^{th}$ term of an AP is given as $a_n = a+(n-1)d$ Therefore $7=a+(7-1)(-4)$ $7=a-24$ $a= 28$ Ans: First term of the given AP is 282) Find the sum of first 10 multiples of 3. Solution The first and last term of the multiples of 3 are $3$ and $30$ Sum of $n^{th}$ term of AP is given as $S_n= n/2(a+l)$ =$10/2(3+30)$ =$5× 33$ = $165$3)How many two digits numbers are divisible be 3. Solution Two digits number divisble by 3 are 12,15,.....99 . Let $n$ be the number of two digits numbers divisble by 3. Therefore $a_n=a+(n-1)d$ $99=12+(n-1)3$ $87=(n-1)3$ $29=(n-1)$ $n=30$4) The sum of four consecutive numbers in an AP is 32 and the ratio of the product of first and last term to the product of middle terms is 7:15 . Find the numbers. Solution Let the four consectuve number be $a-3d$ ,$a-d$ ,$a+d$ , $a+3d$ Therefore sum of number is $(a-3d)+(a-d)+(a+d)+(a+3d) = 32$ $4a=32$ $a=8$ Also ratio of the product of first and last term to the product of middle terms is 7:15 = ${(a-3d)(a+3d)}/{(a-d)(a+d)} = 7/15 $ = $(a^2-9d^2)/{(a^2-d^2)} = 7/15$ =$(15a^2-135d^2)= (7a^2-7d^2)$ =$8a^2=128d^2$ =>$a^2=16d^2$ Since $a$= 8 therfore, =>$64=16d^2$ =>$d^2=4$ =>$d = ± 2$ If $d$ = +2, then numbers are 2,6,10,14 If $d$= -2 , then numbers are 14,10,6,25) The sum of three consecutive numbers in an AP is 27 and the ratio of the product of first and last term to the square of middle term is 8:9 . Find the numbers. Solution Let the first 3 terms of the Ap be $(a-d)$ , $a$ and $(a+d)$ Given $(a-d)+a+(a+d) = 27$ $3a=27$ $a=9$ Also given ${(a-d)(a+d)}/a^2= 8/9$ $(a^2-d^2)/a^2 = 8/9$ $(9a^2-9d^2) = 8a^2$ $a^2=9d^2$ Since $a$ = 9, therefore => $d^2=9$ d= ±3 Therefore the series are $6,9,12$ or $12,9,6$6) Which term of AP 3,15,27,39 .. will be 120 more than its 21$^{st}$ term. Solution Common difference of the series is $d=$ $(15-3)=12$, $(27-15)=12$ Let the $n^{th}$ term be 120 more than its 21$^{st}$ term. $t_n=120+ t_21$ $3+(n-1)12= 120 + 3+(21-1)12$ $(n-1)12= 360 $ $(n-1)= 360/12 $ $(n-1)= 30 $ $n=31$ Ans: $31^{st}$ term of the AP will be 120 more than its 21$^{st}$ term7) The $31^{st}$ term of the AP is 120 more than its 21$^{st}$ term. If the 1st term of the AP is 3, find its 51st term. Solution Given $t_31=120+ t_21$ Let the first term of the AP be $a$ and the common difference be $d$. Therefore, $a+(31-1)d=120 + a+(21-1)d$ $a+30d=120+a+20d$ $10d=120$ $d=12$ Therefore $t_{51} = a+(51-1)d$ = $3+ 50×12$ = $603$ Ans:51st term of the AP is 6038) If $S$, the sum of first $n$ terms of an AP is given by $S_n=3n^2-4n$,find the $n$ term. Solution Sum of $nth$ term of AP is $S_n=3n^2-4n$ Therefore the sum of $(n-1)$ term of AP will be = $S_(n-1)= 3(n-1)^2-4(n-1)$ = $3(n^2-2n+1)-4n+4$ =$3n^2 -6n +3 -4n+4$ =$3n^2-10n+7$ Hence $n$th term of the AP is = $S_n-S_{n-1}$ = $3n^2-4n - 3n^2+10n-7$ = $6n-7$9) If the sum of first four term of an AP is 40 and that of first 14 terms is 280. Find the sum of its first $n$ terms. Solution $S_n=n/2(2a+(n-1)d)$ $s_4 = 4/2(2a+(4-1)d)$ $40 = 2(2a+3d)$ $20 = (2a+3d)$ ...........(1) $s_14 = 14/2(2a+(14-1)d)$ $280 = 7(2a+13d)$ $40= (2a+13d)$ ............(2) From (1) and (2) $20=13d-3d$ $20=10d$ $d=2$ and $20=2a+3×2$ $2a=14$ $a=7$ Sum of first $n$ terms of AP = $n/2(2a+(n-1)d)$ = $n/2(2×7+(n-1)2)$ = $n/2(14+2n-2)$ = $n/2(12+2n)$ = $n(6+n)$ = $n^2+6n$10) If $S$, the sum of first $n$ terms of an AP is given by $S_n=n^2+6n+1$,find the $n$ term. Solution Sum of first $n$ term is $S_n=n^2+6n+1$ Sum of $(n-1)th$ term is $S_(n-1)= (n-1)^2+6(n-1)+1$ = $n^2-2n+1+6n-6+1$ = $n^2+4n-4$ Therefore the $nth$ term is = $S_n-S_(n-1)$ = $n^2+6n+1 - n^2-4n+4$ = $2n+5$11) For what value of $k$ will $k+9$,$2k-1$ and $2k+7$ are the consecutive term of an A.P. Solution Since the numbers are consecutive, therefore the common difference is same. i.e $(2k-1)-(k+9)=(2k+7)-(2k-1)$ $(2k-1)+(2k-1)=(2k+7)+(k+9)$ $4k-2=3k+16$ $k=18$12) For A.P show that $a_p+a_{p+2q}=2a_{p+q}$ Solution Let $a$ be the first term and $d$ be the common difference. $a_p+a_{p+2q}$ = $a+(p-1)d + a+(p+2q-1)d$ = $2a+(2p+2q-2)d$ = $2(a+(p+q-1)d)$ = $2a_{p+q}$ Hence proved.13) If $S_n$ denotes the sum of $n$ terms of an A.P whose common difference is $d$ and first term is $a$, find $S_n - 2S_{n-1} + S_{n-2}$ Solution Sum of first $n$ term be $S_n$ ,first $n-1$ term be $S_{n-1}$ and first $n-2$ term be $S_{n-2}$ Therefore $t_n$ term is =$S_n - S_{n-1}$ and $t_{n-1}$ = $S_{n-1}-S_{n-2}$ Therefore $S_n - 2S_{n-1} + S_{n-2}$ = $(S_n - S_{n-1})-(S_{n-1}-S_{n-2})$ = $t_n - t_{n-1}$ = $d$ Ans: $d$