Question

The 7th term of an AP is 39 and 17th term is 69. Find the series

Answer 1

Appropriate Question:

The 7th term of an AP is 39 and it's 17th term is 69. Find the progression.

Solution:

We know that any general term of an Arithmetic Progression (AP) is given by,

• an = a + (n-1)d

Here,

• an = nth term
• a = First term
• d = Common difference
• n = Number of terms

By using this general form, 7th term of the AP will be:

⇒ a7 = a + 6d

⇒ 39 = a + 6d -------Equation(1.)

Similarly, 17th term of the AP will be:

⇒ a17 = a + 16d

⇒ 69 = a + 16d ---------Equation(2.)

Substracting eq.(1) from eq.(2)

⇒ 69 - 39 = a + 16d - (a + 6d)

⇒ 30 = a + 16d - a - 6d

⇒ 30 = 10d

⇒ 30/10 = d

⇒ 3 = d

Therefore the common difference of AP is 3.

Now put d = 3 in equation(1) to find value of first term.

⇒ 39 = a + 6d

⇒ 39 = a + 6(3)

⇒ 39 = a + 18

⇒ 39 - 18 = a

⇒ 21 = a

Therefore the first term of the AP is 21.

Every arithmetic progression is of the form - a, a + d, a+2d, a+3d, . . .

Therefore, the required progression is:-

21, 24, 27, . . .

$\rule{280}{1}$

Additional Information :-

Whenever we are given any two terms of AP, then the common difference of that AP will be given by,

• $\boxed{ \sf d = \frac{a_n - a_k}{n - k} }$

[Where an and ak are two distinct terms of AP]

Let's see the application of this formula!

Question: 10th term and 15th term of an AP are 60 and 80 respectively, find the common difference.

Solution:

Let an = 80 and ak = 60 implies that n = 15 and k = 10.

$\longrightarrow\sf d = \dfrac{a_n - a_k}{n - k}$

$\longrightarrow\sf d = \dfrac{a_{15} - a_{10}}{15 - 10}$

$\longrightarrow\sf d = \dfrac{80 - 60}{5}$

$\longrightarrow\sf d = \dfrac{20}{5}$

$\longrightarrow\sf d = 4$

So the common difference is 4.

Answer 2

Answer:

Using the prescribed formula of AP :

Taking first term or general term (a) same as last term= 184

Taking common term (d) = -3 because when we find terms in reverse or from last.

t7 = a + ( n - 1)d

= 184 + (7–1)×(-3)

= 184 +(6×-3)

= 184 - 18

= 166.