wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

x1,x2,x3, ________ are in A.P. If x1+x7+x10=6 and x3+x8+x12=11, then x3+x8+x22= __________.

A
21
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
15
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
18
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
31
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 21
Let the first term be a and common difference be d
x1+x7+x10=6
(a)+(a+6d)+(a+9d)=6
3a+15d=6eqn(1)

x3+x8+x12=11
(a+2d)+(a+7d)+(a+11d)=11
3a+20d=11eqn(2)
Solving eqn (1) and (2), we get a=3 and d=1

Now, x3+x8+x22=3a+30d=21
So, option (A)

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Arithmetic Progression
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon