# Find out the first term and the ratio in a arithmetic progression. We know the 3rd term = 7 and the 7th term of the progression= 8

pohnpei397 | Certified Educator

We can see that the interval between each of the terms of this arithmetic progression must be .25.  We know that the interval between terms of this kind of progression must always be the same.  We know that there are three terms between 7 and 8.  That means that term 4 will be 7.25, term 5 will be 7.5 and term 6 will be 7.75.

So, if 7 was the third term, there were two terms before it.  Each was .25 smaller than the one before it.  Therefore, term 2 was 6.75 and term 1 was 6.5.

If you need to show work, the formula for this is

An = A1 + (N + 1) D

Where N is the number of the term and D is the ratio.

So in this case

3 = A1 + (3-1) D

and

7 = A1 + (7-1) D

hala718 | Certified Educator

Let us assume that the arthimitic progression is:

n1, n2, n3, n4, n5, n6, n7

We know that n3=7 and n7=8

Assume that r is the ratio between terms:

==> n3 = n1+(3-1)r =7 ==> n1+2r=7...(1)

==> n7= n1+(7-1)r=8 ==> n1+6r=8...(2)

Multiply (1) with -3 and add t0 (2)

==> -2n1 = -13

==> n1= 13/2= 6.5

giorgiana1976 | Student

The formula for any term of an arithmetic progression is:

an=a1 + (n-1)r, where a1 is the first term and r is the ratio.

a3=a1 + (3-1)r

a7=a1 + (7-1)r

We'll perform the substitutions where it is possible, so that we'll get:

8 = a1 + 2r

20 = a1 + 6r

We are subtracting the second relation from the first one and the result will be

8-20 = a1 + 2r -a1 - 6r

-12 = -4r

r=3

With the found value of r, we go back in the first relation and making substitution of r.

8 = a1 + 2r

8 = a1 + 2x3

8 = a1 + 6

a1= 8-6

a1=2

krishna-agrawala | Student

An arithmetic progression is a series in which the difference between any term and the next one is constant. Thus an arithmetic progression has the following general form:

a, a+b, a+2b, a+3b, a+4b, a+5b, a+6b, ....

Where a and b are constants. Thus the term at nth position in the series is given by:

It is given that the 3rd term of the given series is equal to 7. Therefore:

a + (3-1) = 7

a + 2b = 7   ...   (1)

Similarly the 7rd term of the given series is equal to 7. Therefore:

a + (7-1) = 8

a + 6b = 8   ...   (2)

Subtracting equation (1) from equation (2) we get:

a - a + 6b - 2b = 8 - 7

4b = 1

Therefore:

b = 1/4 = 0.25

Substituting this value of b in equation (1) we get

a + 2*0.25 = 7

a = 7 - 1*0.25 = 7 - 0.5 = 6.5

The first term of the progression is equal to a.

Therefore value of first term of progression is 6.5

The difference between any term and the next term = b = 0.25.

Please note that in an arithmetic progression there is no fixed ratios for the terms in series.

neela | Student

(Note: In arithmetic progression, the suucessive yerms have a common difference and not common ratio.The term common ratio is the ration in Geometric progression)

nth term Tn  = a +(n-1)d, where a is the 1st term and d is the common difference.

T7 -T3 = 8-7 = 1 = a+(7-1)d - [a+(3-1)d] = 4d. Or

d = 1/4.

Also T3 = 7 = a+(3-1)d = a+2*(1/4) . Or a = 7-2*(1/4) = 6and 1/2= 6.5.

So the first term = a = 6.5 and the common difference is 1/4 0r 0.25.