Question

If the eight terms of an ap is 31 and the fifteenth term is 16 more than the eleventh term, Find the Ap
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Answer 1

Answer:

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Step-by-step explanation:

Let the first term of the Arithmetic Progression be a1 , second term be a2 , and so on.

Then the 8th term will be a8=31

Recall the formula to find the last term, or an ,

an=a+(n−1)∗d , where an is the last term, a is the first term, n is the number of terms, and d is the common difference.

Therefore, the above equation can be written as:

a+(8−1)∗d=31

or, a+7d=31 ……………………….[1]

Neither a nor d are known yet, so we’ll leave it there until we find any one of them. Mark it as [1].

Now, it is given that the 15th term is 16 more than the 11th term.

Therefore, a+(15−1)∗d=16+(a+(11−1)∗d))

a+14d=16+a+10d

Cancelling a from both sides, we get —

14d=16+10d

14d−10d=16

4d=16

d=164

d=4

Now that we know our common difference d , we can substitute it in [1].

a+7∗4=31

a+28=31

a=31−28

a=3

Therefore, our A.P. is —

3,(3+4),(3+4+4)...

or simply,

3,7,11,15.....

Answer 2

Answer:

if a is the 1st term and d is the common difference then

a+7d=31

a+14d=16 +a +10d

a+14d-a-10d=16

4d =16

d=4

so ,a+7×4=31

a+28=31

a=31-28=3

then tha Ap series is 3,7,11,15....