# Let T_r be the r-th term of an A.P. for 1, 2, 3, .... If T_3=18 and T_6=39, find T_18. 132

## Question

Let $T_{r}$ be the $r^{\text{th}}$ term of an A.P. for $1, 2, 3, \ldots$. If $T_{3}=18$ and $T_{6}=39$, find $T_{18}$. 132

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## Answer

To find the $18^{\text{th}}$ term ($T_{18}$) of the arithmetic progression (A.P.), we begin by using the given information:

The general form for the $r^{\text{th}}$ term of an A.P. is given by: $$ T_r = a + (r-1)d $$

Given:

- $T_3 = 18$
- $T_6 = 39$

First, let's use these terms to find the first term ($a$) and the common difference ($d$).

For the $6^{\text{th}}$ term: $$ T_6 = a + 5d $$

For the $3^{\text{th}}$ term: $$ T_3 = a + 2d $$

Setting up the equations and solving for $d$: $$ \begin{array}{l} T_6 - T_3 = (a + 5d) - (a + 2d) \ 39 - 18 = 3d \ 21 = 3d \ d = 7 \end{array} $$

Next, we substitute $d = 7$ back into one of the equations to find $a$. Using $T_3 = 18$: $$ 18 = a + 2(7) $$ $$ 18 = a + 14 $$ $$ a = 4 $$

Now, we can find $T_{18}$: $$ T_{18} = a + 17d $$ $$ T_{18} = 4 + 17(7) $$ $$ T_{18} = 4 + 119 $$ $$ T_{18} = 123 $$

Thus, we have the $18^{\text{th}}$ term of the arithmetic progression: $$ \boxed{123} $$

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