Hello all, Here is a problem on Arithmetic progressions. Please help me in solving this. 1) The sum of n(n+2) terms of an A.P with the first term as a and common difference as d is 0. If the series is extended backwards by m terms (i.e., if a-d, a-2d, a-3d, ... a-(m-1)d are also included) and a-(m-1)d is designated as the first term, then what is the sum of n^2+2n+2m terms of the series? Also there is another problem which I found difficult to solve. This is not on progressions. 2) D1, D2, D3 & D4 are recurring decimals given by D1 = 0.q1q2q3q4 ( all 4 decimals recurring) Where q1, q2, q3, q4 are any 4 digits from 1 to 9. Which of the following numbers when multiplied by at least one of the above mentioned decimal will always result in an integer? Thanks in advance. |
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hi..
dis z da kind f questnz u cnt solve it frm frwrd approach...even if ya r able 2 bt dis wnt help ya in savin tym in CAT...so i suggest ya to strt solvin it frm bckwrd aproach....optinz vil also help ya in dis questn...
strt solvin it lyk...
let n=1
thn total no f terms vil b=3 by solvin n*(n+2)
ie -1,0,1 vil b the terms
nw tke m=2(hypotheticaly)
n strt solvin it wot questn says.......
as i said you have to take bckwrd approach n help frm qptions also...as i dnt hv options i request ya to solve the options by puttin ma values...nw herz da soln:
let n=1
thn totl no f terms becum 1(1+2)=3
n the othr cndn z to make sum =0, this v cn make by tkin termz -1,0,1 or -2,0,2 or -3,0,3.......digits wnt make ne diff...so tke -1,0,1(suggested) after dat tke 2nd cnditin..which z to extendin the series bck by m termz..such that a-(m-1)d z designated as first term...so take m=2, series becumz..0,-1,-2,-3,-4
nw put values of m and n in the eqn given in the questn "x=n^2+2n+2m" thn x=7
nw cross check it vid the options given by puttin values of n and m..the option which gives value '7' is da crrect 1...this z ma pt f view...
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