Question

how many terms of Ap series 2,6,10,14 ....must be added to get total of 4232?​

Answer 1

Step-by-step explanation:

Answer

(a) 6,10,14,....

Since t

2

−t

1

=t

3

−t

2

Common difference, d=10−6=4

First term a=6

S

n

=

2

n

2a+(n−1)d=

2

n

2×6+(n−1)4=

2

n

12+4n−4=

2

n

4n+8

=n(2n+4)

2n

2

+4n

(b) For the sum to be 240 from the beginning

240=n

2

+4n

⇒240=n(2n+4)

⇒240=n(2n+4)

⇒120=n(n+2)

⇒n

2

+2n−120=0

⇒n

2

+12n−10n−120=0

⇒n(n+12)−10(n+12)=0

⇒(n+12)(n−10)=0

∴ n=−12,10

10 terms should be added to make the sum to be 240

(c) To make the sum to be 240, 'n' should be natural number.

250=n(2n+4)

⇒125=n(n+2)

⇒n

2

+2n−125=0

⇒n

2

+n−125=0

n=

2

−1

1−4×125

Since n is not a natural number, therefore the term 250 cannot exist in the series.

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