Question

The 3rd term=5 and 7th term=9 find AP

Answer 1

LET THE A. P. HAS FIRST TERM a AND COMMON DIFFERENCE d.

NOW A.T.Q.

a3=5 (THIRD TERM OF AP)

a+2d=5 (AS nTH TERM OF AN AP IS a+(n-1)d ) EQUATION 1

SIMILAR;Y a7= 9

a+6d=9 (EQUATION 2)

ON SOLVING EQ. 1 NAD 2

WE GET

a= 3

d= 1

SO AP IS

3,4,5,6......

THUMBS UP PLZ..

HOPE THIS WILL HELP

second method

a3 = 5

therefore, 5=a+2d --------------------(1)

a7=9

therefore, 9=a+8d ---------------------(2)

By subtracting 1 from 2

9=a+6d

5=a+2d

4=4d

d=1

put the value of d in any eqn

5=a+2(1)

a=5-2

=3

Therefore,, the AP is...... 3,4,5,6,7,8,9....

third method-

a3= 5

a7= 9

an= a + (n-1)d

When an= 5, n will be 3.

5 = a + (3-1)d

5 = a + 2d ......eq 1

When an= 9 then n = 7

9 = a + (7-1)d

9 = a + 6d.............. eq2

eq 2 - eq 1

a + 6d = 9

a + 2d = 5

a - a + 6d - 2d = 9 -5

6d - 2d = 4

4d = 4

d = 4/4

d = 1

Putting the value of d in eq 1 ( You can put the value of d in any eq)

5 = a+2d

5 = a + 2(1)

5 = a +2

5 - 2 = a

3 = a

The standard form of an ap is

a, a+d, a+2d, a+3d

3, 3+1, 3+2(1), 3+3(1)

3,4,5,6..... is the AP.

Thanks !

Answer 2

$\large\underline\pink{Given:-}$

• 3rd term of Ap = 5
• 7th term of Ap = 9

$\large\underline\pink{To find:-}$

• Fine the Ap ....?

$\large\underline\pink{Solutions:-}$

an = a + (n - 1)d

• an = nth terms
• a = first term
• n = number of term
• d = common difference

• 3rd term of Ap = 5

⟹ a5 = a + (3 - 1) d

⟹ 5 = a + 2d ________(i).

• 7th term of Ap = 9

⟹ a9 = a + (7 - 1) d

⟹ 9 = a + 6d _______(ii).

Now, Solving the Eq. (ii) and (i). we get,

${a} \: + \: {6d} \: \: = \: \: {9} \\ {a} \: + \: {2d} \: \: = \: \: {5} \\ \underline{- \: \: \: \: \: \: \: \: \: - \: \: \: = \: \: \: - \: \: \: \: \: } \\ \: \: \: \: \: \: \: {-4d} \: \: \: = \: \: {-4}$

$\: \: \: \: \: \leadsto \: \: d \: \: = \: \: \cancel{\frac{-4}{-4}}$

$\: \: \: \: \: \leadsto \: \: d \: \: = \: \: {1}$

Now, putting the value of d in Eq. (i)

⟹ a + 2d = 5

⟹ a + 2 × 1 = 5

⟹ a + 2 = 5

⟹ a = 5 - 2

⟹ a = 3

Now, the Ap a, a + d and a + 2d

• a = 3
• a + d = 3 + 1 = 4
• a + 2d = 3 + 2(1) = 3 + 2 = 5

Hence, Ap is 3, 4, 5 ..........