Question

Find tn for following A.P. 3,8,13,18.......and then find 30th term of A.P.

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore t_{n}=3n-2}}$

${\bold{\therefore a_{30}=148}}$

${\bold{\underline{\underline{Step\:by\:step\:explanation:}}}}$

• In the given question information given about an A.P whose first term and common difference is given.

• We have to find the tn term and 30th term of this A.P.

$\underline \bold{Given : } \\ \implies A.P = 3,8,13,18,.... \\ \\ \implies First \: term (a)= 3 \\ \\ \implies Common \: difference(d) = 5 \\ \\\implies n=30\\\\ \underline \bold{To \: Find : } \\ \implies t_{n} = ? \\ \\ \implies a_{30} = ?$

• According to given question :

$\bold{Using \: formula \: of \: n_{th} \: term: } \\ \implies a_{n} = a +( n - 1)d \\ \\ \implies a_{n} = 3 + (n - 1) \times 5 \\ \\ \implies a_{n} = 3 + 3n {- 5} \\ \\ \bold{\implies a_{n} = 3n-2} \\ \\ \bold{For \: finding \: a_{30} \: term } \\ \implies a_{n} = a +( n - 1)d \\ \\ \implies a_{30} = 3 + (30 - 1) \times 5 \\ \\ \implies a_{30} = 3 + 29 \times 5 \\ \\ \implies a_{30} = 3 + 145 \\ \\ \bold{\implies a_{30} = 148}$

Answer 2

Solution:

Given:

=> A.P = 3, 8, 13, 18.....

=> a = 3

=> d = $\sf{a_{2}-a}$

       = 8 - 3

       = 5

To Find:

=> 30th term of A.P

Formula used:

$\sf{\implies T_{n}=a+(n-1)d}$

Firstly we calculate Tn,

$\sf{\implies T_{n}=a+(n-1)d}$

$\sf{\implies T_{n}=3+(n-1)5}$

$\sf{\implies T_{n}=3+5n-5}$

$\sf{\implies T_{n}=3n-2}$

Now, we will find 30th term

$\sf{\implies T_{n}=a+(n-1)d}$

$\sf{\implies T_{30}=3+(30-1)5}$

$\sf{\implies T_{30}=3+29\times 5}$

$\sf{\implies T_{30}=3+145}$

$\bf{\implies T_{30}=148}$

Hence, the 30th term is 148.