Question

5th term is 14 and 12th term is 35 find ap

Answer 1

Answer:

t5=14

a+(5-1)d=14

a+4d=14

again,

t12=35

a+(12-1)d=35

a+11d=35

Now by eliminating method (you can use any method for solving equations)

a+4d=14

-(a+11d=35)

then -7d=-21

d=3

then a+4d=14

again a+4(3)=14

a=14-12

a=2

then the A.P is as below

2,5,8,11,14..........

Answer 2

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

• 5th term of Ap = 14
• 12th term of Ap = 35

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

• 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

5th term of Ap = 14

⟹ a14 = a + (5 - 1) d

⟹ 14 = a + 4d ________(i).

12th term of Ap = 35

⟹ a35 = a + (12 - 1) d

⟹ 35 = a + 11d _______(ii).

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

${a} \: + \: {11d} \: \: = \: \: {35} \\ {a} \: + \: {4d} \: \: = \: \: {14} \\ \underline{\: \: \: - \: \: \: \: \: \: - \: \: \: \: \: = \: \: \: \: \: - \: \: \: \: \: } \\ \: \: \: \: \: \: \: \: \: \: \: {7d} \: \: \: \: \: = \: \: \: {21}$

$\: \: \: \: \: \leadsto \: \: d \: \: = \: \: \frac{21}{7}$

$\: \: \: \: \: \leadsto \: \: d \: \: = \: \: {3}$

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

⟹ a + 4d = 14

⟹ a + 4 × 3 = 14

⟹ a + 12 = 14

⟹ a = 14 - 12

⟹ a = 2

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

• a = 2
• a + d = 2 + 3 = 5
• a + 2d = 2 + 2(3) = 2 + 6 = 8

Hence, Ap is 2, 5, 8 ..........