Let
d = the common difference
t(n)=nth term
t(6)-t(5)=d
t(5)-t(4)=d
t(4)-t(3)=d
Add
t(6)-t(3)=3d
or
3d=12
Solve for the common difference d=4
Sum of the 2nd and 5th terms = 42
t(2)+t(5)
=t(2)+(t(2)+3d)
=2*t(2)+12 = 42
Solve for t(2) = 15
t(1) should be t(2)-1=15-4=11
and
t(20)=t(1)+19d
THE SUM OF THE 2ND AND 5TH TERMS OF AN A.P IS 42.
IF THE DIFFERENCE BETWEEN THE 6TH AND 3RD TERM IS 12,
FIND THE:
COMMON DIFFERENCE,
1ST TERM,
20TH TERM.
9 answers
Explain in a simpler way please
I didn't understand
Illustrate
I don't understand
After the step where you expande t5 i just got lost
Understood
Thanks alot
Thanks alot
Please break it down to ss1 level of understanding
please break it down into a more simple form.