CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The product of the third by the sixth term of an arithmetic progression is 406. The division of the ninth term of the progression by the fourth term gives a quotient 2 and a remainder 6. Find the first term and the difference of the progression.

Open in App
Solution

Let a be the first term and d be the common difference and we know that nth term, Tn=a+(n1)d
T3=a+2d&T6=a+5dT3×T6=406
(a+2d)(a+5d)=406
a2+2ad+5ad+10a2=406
a2+7ad+10d2=406(i)Now,T9=T4×2+6
a+8d=(a+3d)2+6
a+8d=2a+6d+6
a2d+6=0
=2d6(ii)
Replacing a from (ii) in (i).
(2d6)2+7(2d6)d+10d2=406
4d224d+36+14d242d+10d2=406
28d266d+36=406
14d233d+18=203
14d233d185=0
Using quadratic equation:
d=(33)±(33)24×14×(185)2×14=33±(33)2+4×14×18528=33±1089+1036028=33±10728d=33+10728,3310728=14028,7428=5,3728
But common difference cannot be negative in this case.
Common difference=5.
From (ii)
a=2d6=2×56=106=4
The first term is 4.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
General Form of an AP
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon