What is the first term of an arithmetic progression of positive integers ?

(I) Sum of the squares of the first and the second term is 116.

(II)The fifth term is divisible by 7.

(a) A (b) B (c) C (d) D

• sum of ^2 of 1st and 2nd number is 116.
  =>there is only possibility is (4)^2+(10)^2=116 ' . ' [16+100=116]
  =>here D( difference ) is 6.
  => 5th term of this Ap is 28 i.e[4+10+16+22+28............... ]
  here only satisfy the above condition i.e 5th term is devided by 7
  So as for question 1st term of this Ap is a=4.
  
• 9 years agoHelpfull: Yes(20) No(2)
• It is given that sum of squares of first and second term is 116
  therefore, a^2 + b^2 =116
  Now a and b can have many values such as
  for a=10,b=4 | for a=9,b=5
  and it is not mention what is the common difference of AP
  so we can take AP of any common difference
  Hence either first term is 4 or 5
  and forming AP with first term 4 and 2nd term 10 we can see that 5th term is divisible by 7 same is true for AP with first term 5 and common difference 4.
  Hence we can find the first term using statement 1 only but statement 2 alone can't be helpful.
• 9 years agoHelpfull: Yes(17) No(20)
• @ABDULLAH....how you can take a=9 and b=5, becoz the square sum of both is not equal to 116
• 9 years agoHelpfull: Yes(9) No(2)
• Sorry @deepak Gupta i was little dumb then to take 9 and 5 but the solution is right with 10 and 4
• 9 years agoHelpfull: Yes(6) No(0)
• (A) It is given in statement (I) that sum of squares of the first and second term is 116
  and there exist two such numbers only 4 and 10 such that (4)2 + (10)2 = 116. Hence question can
  be answered with the help of statement (I) alone.
• 9 years agoHelpfull: Yes(4) No(0)
• we can't say surely what will be the first term of this A.P. it can be 4 or 5........so I think both statements are not enough to answer this question.......
• 9 years agoHelpfull: Yes(3) No(8)
• in an AP a^2+(a+d)^2=116
  where a= first no.
  and d= common difference of ap
  for unit 6 some possible cases are (51,15,24,42,33,06,60),
  but in square case 33 case is not possible
  so first try 51
  then for 5 ........25........(...5)^2
  and for 1............81......(..9)^2
  so we find 81 +25 =116
  5th term is Tn=a+(n-1)d
  then d=4 ,a=5
  so Tn=21
  second condition is also true
  
• 9 years agoHelpfull: Yes(1) No(17)
• 9^2+5^2 =106 not 116 and
  If we take 4 as a first term and 10 as second then 5th term will be 28 and it is divisible by 7 so both statements will be required
• 9 years agoHelpfull: Yes(1) No(0)
• We hav 2 Variable and 1 Soln.so, Use hit and trial method :
  ans : 1st term - 4 , Common diff - 6
• 9 years agoHelpfull: Yes(1) No(0)
• as it is given that sum of squares of first 2 terms is 116 so
  a^2 +b^2=116 so the only condition satisfied is a should be =4 as well b will be =10.
  now this forms an A.P with first term 4 & another term as 10
  AP will be 4+10+16+22+28+34+.....
  clearly 5th term is divisible by 7 hence both statements are satisfied
• 9 years agoHelpfull: Yes(1) No(0)
• both statement are not enough to give the ans
• 9 years agoHelpfull: Yes(0) No(2)
• as both I and II conditions are not giving a unique and definite answer.So, we can say that both the statements are together are not enough to answer this question. D is ans
• 9 years agoHelpfull: Yes(0) No(1)
• here both a^2+b^2=116, so a and b can both be +ve number ex -
  a=10,b=4 or
  a=-10,b=-4 or
  a=-10,b=4 or
  a=10,b=-4 ,so statement 1 is not enough to answer the question
• 9 years agoHelpfull: Yes(0) No(0)
• @ Abdullah how ccan you tell dt we can find the first term using statement 1 only but statement 2 alone can't be helpful? according to me second statement is useful to find the answer exactly since by using first statement alone we are getting more than one answer... so answer should be both statements are required to solve...
  am i correct? please clear my doubt............
  
• 9 years agoHelpfull: Yes(0) No(0)
• with first statement only we can answer .....plz refer elitmuszone.com
  
• 9 years agoHelpfull: Yes(0) No(0)
• answer is a=4
• 7 years agoHelpfull: Yes(0) No(0)