# In an AP, it is given that, S5 + S7 =167 and S10= 235, then find the AP, where Sn denotes the sum of its first n terms.

Let the first term be a.

Let Common difference be d.

Given: S5 + S7 =167

S10= 235

Now, Sum of n terms of an arithmetic series is given by:

Sn = [2a + (n - 1)d]

So, consider

S5 + S7 =167

(5/2) [2a + (5 - 1)d] + (7/2) [2a + (7 - 1)d] = 167

(5/2) [2a + 4d] + (7/2) [2a + 6d] = 167

5 × [a + 2d] + 7 × [a + 3d] = 167

5a + 10d + 7a + 21d = 167

12a + 31d = 167 …………….. (1)

Now, consider S10= 235

(10/2) [2a + (10 - 1)d] = 235

5 × [2a + 9d] = 235

10a + 45d = 235

2a + 9d = 47 ………. (2)

Subtracting equation (1) from 6 times of equation (2), we get,

23d = 115

d = 5

So, from equation (2), we get,

a = 1/2 (47 - 9d)

a = 1/2 (47 - 45)

a = 1/2 (2)

a = 1

Therefore the AP is a, a + d, a + 2d, a + 3d,…

i.e. 1, 6, 11, 16, ….

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Champ Quiz | Arithmetic Progression34 mins
Champ Quiz | Arithmetic Progression30 mins
Lets Check Your Knowledge in A.P.49 mins
Arithmetic progression: Previous Year NTSE Questions34 mins
Quiz on Arithmetic Progression Quiz32 mins
Get to Know About Geometric Progression41 mins
NCERT | Solving Questions on Introduction of A.P42 mins
Arithmetic Progression Tricks and QUIZ37 mins
Quiz | Group of Questions on General Term of an A.P49 mins
Become a Master of A.P. in 45 Minutes!!47 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses