Question

An arithmetic sequence has its 5th term equal to 22 and its 15th term equal to 62 find its 100th term

Answer 1

Answer:

Let the first term of AP is a

and common difference is d

then

ATQ

we get

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.

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a+4d=22 ----->> 1

a+14d=62 ------>> 2

subract eq. 1 from 2

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.

then

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.

10d=40

d=4

and a=6

th term is

a+99d

=6+99×4

=410

ok

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Answer 2

Answer:

t5=22

t15=62

t100=?

Step-by-step explanation:

t5 can be written as a+4d=22...(1)

t15 can be written as a+14d=62.....(2)

t100 can be written as a+99d...(3)

subtracting eqn (2) from (1)

a+4d=22

-a+14d=62

-10d= -40

d= 4

substituting d=4 in eqn (1)

a+4d=22

a+4(4)=22

a+16=22

a=22-16

a=6

substituting a=5 and d=4 in eqn (3)

a+99d

6+99(4)

6+396

410

t10 =410