The 4th and 7th term of an arithmetic progression are 11 and -4 respectively. What is the 15th term ?
A)-44 B)-39 C)-49 D)-34

• 4th term of A.P => a + 3d = 11, 7th term=> a + 6d = -4
  On solving the two equations, we get d = -5
  Substituting d= -5 in one of the equations, we get a = 26
  Now, 15th term of A.P will be,
  a + 14d
  Substituting d=-5 and a=26 we get
  (26) + 14(-5) = -44
  Hence, answer is -44
• 3 years agoHelpfull: Yes(11) No(0)
• answer--> 4th term = a+3d-------(1)
  7th term= a+6d--------(2)
  15th term=a+14d------(3)
  equating both (1) and (2) we get
  d= -5 then put the value of d in equation (1):
  we get a=26 finally put all the values in equation (3)
  a+14d = 26+14(-5) = -44. thank you!
• 3 years agoHelpfull: Yes(4) No(0)
• Finding the equation a + (N-1)d = a_n
  after solving the equation we will get the solution
• 3 years agoHelpfull: Yes(2) No(0)
• a+3d=11;
  a+6d=-4;
  -3d=11+4;
  d=-5;
  a+6*-5=-4;
  a=26;
  15th term=a+14d;
  26+14*-5=-44
  hence 15th term is -44
• 3 years agoHelpfull: Yes(2) No(0)
• a+3d=11
  a+6d=-4
  a=26
  d=-5
  15th term = a+14d
  26+14(-5) = -44
• 3 years agoHelpfull: Yes(1) No(0)
• a+3d=11
  a+6d=-4
  by solving we get d=-5
  now sub d in a+14d we get
  -44
• 3 years agoHelpfull: Yes(1) No(0)
• a4=a+3d=11
  a7=a+6d=-4
  d=-5
  a4=a+3(-5)=11
  a=26
  a15=a+14d=26+14(-5)=26-70=-44
• 3 years agoHelpfull: Yes(1) No(0)
• a+3d=11
  a+6d=-4
  On equating the above equations we get a = 26; d =-(15/3).
  a+14d = 15th term => -44
• 3 years agoHelpfull: Yes(1) No(0)
• 4th term =a+3d=11------1)
  7th term = a+6d=-4------2)
  15th term =a+14d-----3)
  subtract equation 1by2 we get
  3d=-4-
• 3 years agoHelpfull: Yes(1) No(0)
• Using the AP formula Tn = 2a+(n-1)d and solving the equations the answer will be -44
• 3 years agoHelpfull: Yes(0) No(2)
• Ans. -44
  Easy one
• 3 years agoHelpfull: Yes(0) No(0)
• option A is correct
• 3 years agoHelpfull: Yes(0) No(0)
• a+3d=11 -3d=15
  a+6d=-4 d=-5 a+3d=11=>a+3(-5)=11=>a=11+15=26
  - - + 15th term=>a+14d=>26+14(-5)=-44
  -------------------
  -3d=15
• 3 years agoHelpfull: Yes(0) No(0)
• a=26
  d=-5
  a+14d=-44
• 3 years agoHelpfull: Yes(0) No(0)
• This is arithmetic progression
  so, by that I found d(Common diff)=-5 a(First term)=26 and
  a15th= a+14d=-44
  ANS.-44
• 3 years agoHelpfull: Yes(0) No(0)
• Ans:-44
  Explanation: Because the difference of A.P is 15. It has given the term 4th and 7th term that means the A.P of difference of 3 terms is 15. Here it asked us to find the 15th term. 4th term=11,7th term=11-15=-4,10th term=-4-15=-19,13th term=-19-15=-33 and if we can see that each term gets incremented the difference of A.P is 5. So finally 15th term=-33-10=-44.We got the answer..:)
• 3 years agoHelpfull: Yes(0) No(0)
• Given,
  t4 = 11,
  t7 = -4
  as we know that :-
  tn = a+(n-1)d
  So apply this formulla,
  t4= a+(4-1)d
  11 = a+3d--------------- equation (1)
  
  similarary,
  
  t7 = a+(7-1)d
  -4 =a+6d -------------------equation (2)
  
  Equation(2) - Equation(1),
  
  a+6d-(a+3d) = -4-11
  a+6d-a-3d = -15
  3d = -15
  d = -15/3
  d = -3
  put the (d = -3) in equation (1),
  
  then ,
  11 = a+3(-5)
  11 = a-15
  a = 26
  
  Hence,
  t15 = a + (15-1)d
  t15 = 26 + 14*(-5)
  t15 = 26 + (-70)
  t15 = 26 - 70
  
  t15 = -44 answer.
• 3 years agoHelpfull: Yes(0) No(0)
• 4th term = a4 = a + 4d,
  7th term = a7 = a + 6d.
  given, a4 = 11 and a7 = -4,
  11 = a + 4d -------------(i)
  -4 = a + 6d --------------(ii)
  On solving equation i and ii,
  a=26,
  d=-5
  a15 = a + 14d = 26 + 14(-5)
  ANS : a15 = -44
• 3 years agoHelpfull: Yes(0) No(0)
• Area of shaded area=(area of square-area of circle)+area of sector
  =(196-22/7*7*7)+LR/2
  =42+19.25
  =61.25
• 3 years agoHelpfull: Yes(0) No(0)
• on solving, we got an equation like this!
  4 the term= a+3d= 11 (1)
  7th term= a+6d=-4 (2)
  on solving both equation we get
  a=26
  and d=-5
  
  putting this date in 15th term that is :-
  a+14d= 26=(14*-5)=26-70= -44 is the answer.
• 3 years agoHelpfull: Yes(0) No(0)
• Basic formula for AP : T(n)=a+(n-1)d;
  (i) given T(4)=11 ==> a+3d=11
  T(7)=-4 ==> a+6d=-4
  ______-_________________
  -3d=15
  d=-5
  (ii) put the value of d in any equation above and find a:
  a+3(-5)=11
  a=26
  (iii)we have to find T(15):
  T(15)=a+14d
  =26+14*-5
  =(-44) Ans
• 3 years agoHelpfull: Yes(0) No(0)
• 4th term is 11
  5th term 6
  6th term 1
  7th term -4
  
  d=11-6=5
  remaining terms are 8 (8*5=45)
  45 should be in negative , therefore -45-4=-44
• 3 years agoHelpfull: Yes(0) No(1)