1) the 16 term of an ap is 5 times its third term if 10 term is 41 then find the sum of first 15 terms
2)find 60th term of an ap 8 10 12 if it has a total of 60 terms and find the sum of its last 10 them
3)in an ap if S5 Plus S7 equal to 167 and S10 equal to 235 then find the AP SN denotes the sum of first n terms
answer the three questions and earn 20 points
Answers
1) HI....
16th term=5(3rd term) a₁₀=a+9d
a+15d=5(a+2d) 41=5d/4+9d
a+15d=5a+10d 41=41d/4
15d-10d=5a-a 41*4/41=d
5d=4a =>d=4
a=5d/4 a=5(4)/4 =>a=5
Sn=n/2[2a+(n-1)d]
S₁₅=15/2[2(5)+(14)4]
=15/2[10+56]
=15/2[66]
S₁₅=495.
2) Sn=a+n(n-10)d d=a₂-a₁=10-8=2
S₆₀=8+(59)2
=8+118
S₆₀=126.
To find the sum of the last 10 term , i just found out the 50th term(because 60-10=50) and substituted it in the formula Sn=n/2(a+l).
a₅₀=a+49d
=8+(49)2
=8+98
a₅₀=106
S(last 10 terms)=n/2[a+l}
=10/2[106+126]
=5[232]
.: S(last 10 terms)=1160.
3) S₅+S₇=167 Using the formula Sn=n/2[2a+(n-1)d]
We get , 5/2[2a+4d]+{7/2[2a+6d]}=167
5a+10d+7a+21d=167
12a+31d=167 (eq.1)
It is given S₁₀=235
10/2[2a+9d]=235
10a+45d=235 (eq.2)
Usind elimination method for eq.1 and eq.2
We get d=5.
12a+31d=167
12a=167-31(5)
12a=167-155
12a=12
=> a=1.
Now that we have got a and d, we can find the A.P
.:The A.P is 1, 6, 11, 16.....
HOPE IT WAS HELPFUL....!