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Which term of the A.P 5, 12, 19, 26 .................... is 145?$\left( a \right)$ 12$\left( b \right)$ 18$\left( c \right)$ 25$\left( d \right)$ 21

Last updated date: 21st Jun 2024
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Hint: In this particular question use the concept that the last term of an A.P is given as, ${a_n} = {a_1} + \left( {n - 1} \right)d$, where symbols have their usual meanings so use these concepts to reach the solution of the question.

Given A.P
5, 12, 19, 26 .............
Now we have to find out which term is 145.
Now as we know that last term or ${n^{th}}$ term of an A.P is given as, ${a_n} = {a_1} + \left( {n - 1} \right)d$................. (1)
Where, ${a_n}$ is the last term, ${a_1}$ is the first term, d is the common difference and n is the number of terms in an A.P.
Let 145 be the last term of this given A.P, therefore, ${a_n} = 145$.
Now in the given A.P first term is, ${a_1}$ = 5.
Common difference (d) = (12 – 5) = (19 – 12) = 7
Now substitute these values in equation (1) we have,
$\Rightarrow 145 = 5 + \left( {n - 1} \right)7$
Now simplify this we have,
$\Rightarrow 145 - 5 = \left( {n - 1} \right)7$
$\Rightarrow \left( {n - 1} \right)7 = 140$
$\Rightarrow \left( {n - 1} \right) = 20$
$\Rightarrow n = 20 + 1 = 21$
So 145 is the ${\left( {21} \right)^{st}}$ term of the given A.P.
Hence option d is the correct answer.

Note:Whenever we face such types of questions the key concept we have to remember is that always recall all the basic formulas of an A.P from which one of them is stated above so simply substitute the values in this formula as above and simplify we will get the required answer.