Latest TCS Aptitude Question SOLUTION: 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

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Arithmetic

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40167065) TCS Numerical Ability question Keep an EYE

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

Asked In TCS Devendra Marghade (3 years ago)

Unsolved Read Solution (22)

Is this Puzzle helpful?   (152)   (30) Submit Your Solution Arithmetic

Read Solution (22) :

Ayushi  (3 years ago)

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

Is this solution Helpfull? Yes (11) | No   

Aditya Ashutosh  (3 years ago)

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!

Is this solution Helpfull? Yes (4) | No   

Sanjeev Yadav  (3 years ago)

Finding the equation a + (N-1)d = a_n
after solving the equation we will get the solution

Is this solution Helpfull? Yes (2) | No   

Naren Surya Tanuj Dasari  (3 years ago)

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

Is this solution Helpfull? Yes (2) | No   

Abhilash  (3 years ago)

a+3d=11
a+6d=-4
a=26
d=-5
15th term = a+14d
26+14(-5) = -44

Is this solution Helpfull? Yes (1) | No   

vissu  (3 years ago)

a+3d=11
a+6d=-4
by solving we get d=-5
now sub d in a+14d we get
-44

Is this solution Helpfull? Yes (1) | No   

ujwala  (2 years ago)

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

Is this solution Helpfull? Yes (1) | No   

Prasanna Kumar  (2 years ago)

a+3d=11
a+6d=-4
On equating the above equations we get a = 26; d =-(15/3).
a+14d = 15th term => -44

Is this solution Helpfull? Yes (1) | No   

Madhav Sanjay Bagal  (2 years ago)

4th term =a+3d=11------1)
7th term = a+6d=-4------2)
15th term =a+14d-----3)
subtract equation 1by2 we get
3d=-4-

Is this solution Helpfull? Yes (1) | No   

Vaibhav Raj Srivastava  (3 years ago)

Using the AP formula Tn = 2a+(n-1)d and solving the equations the answer will be -44

Is this solution Helpfull? Yes | No (2)  

Suman Kumar  (3 years ago)

Ans. -44
Easy one

Is this solution Helpfull? Yes | No   

Sakshi Kumari  (3 years ago)

option A is correct

Is this solution Helpfull? Yes | No   

sharanya  (2 years ago)

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

Is this solution Helpfull? Yes | No   

sai sailaja  (2 years ago)

a=26
d=-5
a+14d=-44

Is this solution Helpfull? Yes | No   

Vaidehi Mhaiskar  (2 years ago)

This is arithmetic progression
so, by that I found d(Common diff)=-5 a(First term)=26 and
a15th= a+14d=-44
ANS.-44

Is this solution Helpfull? Yes | No   

KUMARAN L S  (2 years ago)

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..:)

Is this solution Helpfull? Yes | No   

SUMIT RAJ  (2 years ago)

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.

Is this solution Helpfull? Yes | No   

prabhu  (2 years ago)

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

Is this solution Helpfull? Yes | No   

Seelam Pranavi  (2 years ago)

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

Is this solution Helpfull? Yes | No   

karan kanu  (2 years ago)

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.

Is this solution Helpfull? Yes | No   

vansh  (2 years ago)

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

Is this solution Helpfull? Yes | No   

Sahana  (2 years ago)

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

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.