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

• let a be the first term and a+d be 2nd term term and a+4d is 5th term
  it is given that a^2+(a+d)^2=116
  so putting a=4 and d=6 then a+d=4+6=10 so 4^2+10^2=110 and a+4d=4+4*6=28 which is divisible by 7.
  hence first term will be 4.
• 8 years agoHelpfull: Yes(7) No(1)
• the sum of square of two no is 116 i.e no lie 1-11, because 11*11 =121 which is greater than 116
  again after cheaking all possible pair of no only 4 and 10 satisfy this condsn , again the progression is of positive integer so 10 cant not be 1st term
  so the 1st term is 4,
  but from the 2nd we cant able to find the 1st term
  so only 1st statement is sufficient to answer the question
  
• 8 years agoHelpfull: Yes(3) No(1)
• Here if the 1st term is 4 and second term is 10 then only the condition is satisfied and the fifth term is 28(difference of 6 between every no) which is divisible by 7 and here the 1st statement is necessary to answer the question because from the second statement we can't find the 1st term of the sequence.
• 8 years agoHelpfull: Yes(0) No(0)
• Consider the first and second term be 4 and 10
  a=4,d=6,
  fifth term = 28 (divisible by 7)
  Hence the taken terms are correct
  So first term is 4
• 8 years agoHelpfull: Yes(0) No(0)
• A
  1st term 4 and 2nd term is 10
  (4)^2+(10)^2=116
• 7 years agoHelpfull: Yes(0) No(0)
• first statement is sufficient as the cases are being made by using the sum of squares (given) and a comes out to be 4. series is 4,10,....
• 7 years agoHelpfull: Yes(0) No(0)
• Statement1 10sqr+4squr=116
  9sqr+5sqr=116
  Not sufficient
  St2 not sufficient
  Taking both 4 10 16 22 28
  28divd 7 divisible
  5 9 13 17 21
  21 divide 7 divisible
  So option d correct
• 7 years agoHelpfull: Yes(0) No(1)