Question

Determine the AP was third term is 16 and the 7th term exceeds the 5th term by 12

Answer 1

AnswEr :

Given -

• 3rd term is 16 & 7th term exceeds 5th term by 12.

We know the formula,

$\longrightarrow\sf \Big[a_{n} = a + (n - 1) d \Big]$

$\longrightarrow\sf a_{3} = a + (3 - 1)d$

$\longrightarrow\sf a_{3} = a + 2d$

Third term of the AP is 16. So,

$\longrightarrow\sf a + 2d = 16 \;\;\;\;\;\;\: ...eq.(1)$

Similarly,

$\longrightarrow\sf a_{7} = a + 6d \\ \\ \longrightarrow\sf a_{5} = a + 4d$

$\rule{150}2$

$\longrightarrow\sf a_{7} - a_{5} = 12 \\ \\ \longrightarrow\sf a + 6d - [a + 4d] = 12 \:\;\;\; \; a_{7} = a + 6d \: \& \; a_{5} = a + 4d \\ \\ \longrightarrow\sf \cancel{ a} + 6d \: \cancel{- a} - 4d = 12 \\ \\ \longrightarrow\sf 2d = 12 \\ \\ \longrightarrow\sf d = \cancel\dfrac{12}{2} \\ \\ \longrightarrow\sf\red{d = 6}$

Substituting the value of d in eq. (1) -

$\longrightarrow\sf a + 2 \times 6 = 16 \\ \\ \longrightarrow\sf a + 12 = 16 \\ \\ \longrightarrow\sf a = 16 - 12 \\ \\ \longrightarrow\sf\red{a = 4}$

Now, we get the values of a & d.

Hence, the Arithmetic progression is - a, a + d, a + 2d, a + 3d and so on.

$\longrightarrow\sf 4, 4 + 6, 4 + 2 \times 6 , 4 + 3 \times 6 \\ \\ \longrightarrow\sf 4, 10, 16, 22 \: and \; so \; on.$

$\rule{150}2$

Answer 2

Answer:

Let a be the First term,

a3 be the third term,

a5 be the 5th term and

a7 be the 7th term

According to question :-

a3 = 16

a7 = a5 + 12 ............ (1)

Let the common difference be "d"

Common difference is equal in AP

So,

a7 = a5 + d + d = a5 + 2d ............(2)

From Equation (1) & (2)

a5 + 12 = a5 + 2d

2d = 12

d = 6

From Given, we get that

a3 = 16

a3 = a + 2d = 16

a + ( 2 × 6 ) = 16 [ We know that d = 6 ]

a + 12 = 16

a = 4

So first term is 4 .... means a = 4

Now,

We can find AP by adding d continuously

Thus,

AP is 4, 10, 16, 22, 28....... Answer!!!