Question

Find the 110 th term in the sequence 5, 12, 19

Answer 1

GIVEN :–

• Sequence 5 , 12 , 19 ..........

TO FIND :–

• 110th term of of sequence = ?

SOLUTION :–

• Difference of first two terms = 12-5 = 7

• Difference of third and second term = 19-12 = 7

→ Hence , It's an A.P. series.

• We know that nth term of A.P. –

$\\ \large\implies{ \boxed{ \bf T_n = a + (n-1)d}}$

• Here –

$\\ \bf \: \: \: { \huge{.}} \: \: \:First \: \: term(a)=5$

$\\ \bf \: \: \: { \huge{.}} \: \: \:Common \: \: difference(d)=12 - 5 = 7$

$\\ \bf \: \: \: { \huge{.}} \: \: \:n=110$

• Now put the values –

$\\\implies\bf T_n = a + (n-1)d$

$\\\implies\bf T_{110}= 5 + (110-1)(7)$

$\\\implies\bf T_{110} = 5 + (109)(7)$

$\\\implies\bf T_{110}= 5 +763$

$\\\implies \: \: \large{ \boxed{\bf T_{110}= 768}}$

▪︎ Hence , 110th term of A.P. is 768.

Answer 2

Step-by-step explanation:

Given :

• the sequence 5, 12, 19

• Tn = 110

To Find :

• Find the 110 th term

Solution :

D = A2 - A1

Substitute all values :

D = 12 - 5

D = 7

We know that

Tn = a + ( n - 1 ) d

Substitute all values :

T_110 = 5 + ( 110 - 1 ) 7

T_110 = 5 + 109 × 7

T_110 = 5 + 763

T_110 = 768

Hence the T_110 is 768