410. Show that a, a, ..., ..... form an AP where a is defined as below:
(i) a = 3 + 4n
(ii) a = 9-5n
11
Also find the sum of the first 15 terms in each case.
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Answers
Answered by
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Answer:
(i) a
n
=3+4n given
a
1
=3+4(1)=7
a
2
=3+4(2)=11
a
2
=3+4(3)=15
∴d=a
3
−a
1
=11−7=4
Here a=7, d=4 and n=15
By usingS
n
=
2
n
[2a+(n−1)d] we have,
S
15
=
2
15
[2×7+(15−1)4]
=
2
15
(14+56)
=
2
15
×70=525.
Step-by-step explanation:
(ii) a
n
=9−5n given
a
1
=9−5(1)=9−5=4
a
2
=9−5(2)=9−10=−1
a
3
=9−5(3)=9−15=−6
∴d=a
2
−a
1
=−6−(−1)=−5
Here a=4,d=−5 and n=15
By using S
n
=
2
n
[2a+(n−1)d] we have,
S
15
=
2
15
[2×4+(15−1)(−5)]
=
2
15
(8−70)
=
2
15
×(−62)=−465.
This is the answers hope it will help you. ☺️
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