Question

An arithmetic progression has the first term as x and the common difference as d.

a)Write down in terms of x and d, the 3rd, 9th and 25th terms of the progression (1mark)

b)The progression is increasing and the 3rd 9th and 25th terms forms the first three consecutive terms of a geometric series.If the sum of the 7th and twice the 6th term of the arithmetic progression is 78.Calculate the first term and the common difference of the AP

KINDLY DO NOT ATTEMPT IF YOU DO NOT KNOW⛔


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Answer 1

First term: 6

Common difference: 15/4

Explanation:

General form of arithmetic progression:

`a_n=a+(n-1)d`

(where
`a` is the initial term and
`d` is the common difference between terms)

Question (a)

`\implies a_3=a+(3-1)d=a+2d`

`\implies a_9=a+(9-1)d=a+8d`

`\implies a_(25)=a+(25-1)d=a+24d`

Question (b)

General form of a geometric progression:
`a_n=ar^(n-1)`

(where
`a` is the initial term and
`r` is the common ratio)

Therefore, the first three terms of a geometric series are:

`a_1=ar^0=a`

`a_2=ar^1=ar`

`a_3=ar^2`

To find the common ratio:

`r=(ar^2)/(ar)=(ar)/(a) \implies (a_3)/(a_2)=(a_2)/(a_1)`

If the 3rd 9th and 25th terms of an arithmetic progression form the first three consecutive terms of a geometric series, then inputting these into the above formula for r:

`\implies (a+24d)/(a+8d)=(a+8d)/(a+2d)`

`\implies (a+24d)(a+2d)=(a+8d)(a+8d)`

`\implies a^2+26ad+48d^2=a^2+16ad+64d^2`

`\implies 10ad=16d^2`

`\implies 10a=16d`

The 6th and 7th terms of the arithmetic progression are:

`a_6=a+5d`

`a_7=a+6d`

If the sum of the 7th and twice the 6th term of the arithmetic progression is 78, then:

`\implies a_7+2(a_6)=78`

`\implies (a+6d)+2(a+5d)=78`

`\implies 3a+16d=78`

Substituting
`10a=16d` into
`3a+16d=78` and solving for a:

`\implies 3a+10a=78`

`\implies 13a=78`

`\implies a=6`

Substituting
`a=6` into
`3a+16d=78` and solving for d:

`\implies 3(6)+16d=78`

`\implies 16d=60`

`\implies d=(15)/(4)`