find the Sum of the first 10 terms of AP whose 4th and 12terms are 12 and 228 respectively
Answers
Answered by
0
Answer:
a
2
=a+d=14
a+4=14
a=10
Now, sum of 51 terms
S
n
=
2
n
(2a+(n−1)d)
S
51
=
2
51
(2(10)+(51−1)4)
=
2
51
(20+200)
=
2
51×220
=51×110=5610
S
51
=5610
Answered by
0
Answer:
Correct option is A)
a
2
=14 and a
3
=18
Common difference, d=a
3
−a
2
=18−14=4
Now
a
2
=a+d=14
a+4=14
a=10
Now, sum of 51 terms
S
n
=
2
n
(2a+(n−1)d)
S
51
=
2
51
(2(10)+(51−1)4)
=
2
51
(20+200)
=
2
51×220
=51×110=5610
S
51
=5610
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