Question

show that a1,a2,form an Ap where an is defined :an=3+4n​

Answer 1

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

(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