Find the 20th term from the last of the AP:3,8,13,...,253.

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Solution

$\begin{matrix} G i v e n s e r i e s i s 3, 8, 13, . . .253 H e r e, f i r s t t e r m, a = 3 c o m m o n d i f f e r e n c e, d = 5 L e t u s f i n d t h e t o t a l n u m b e r o f t e r m s a t f i r s t . u s e T_{n} = a + (n - 1) d \Rightarrow 253 = 3 + (n - 1) 5 \Rightarrow 250 = (n - 1) 5 \Rightarrow n - 1 = \frac{250}{5} \Rightarrow n - 1 = 50 ∴ n = 51 s o, t h e r e a r e 51 t e r m s i n g i v e n s e r i e s . N o w w e k n o w t h a t m^{t h} t e r m f r o m l a s t = l a s t t e r m - m^{t h} t e r m + 1 s o, 20 t h t e r m f r o m l a s t = 51 - 20 + 1 = 32 H e n c e, 20 t h f o r m l a s t = 32 t h t e r m f r o m f i r s t u s e T_{n} = a + (32 - 1) d = 3 + 31 \times 5 = 3 + 155 = 158 H e n c e, 20 t h t e r m f o r m l a s t = 158. \end{matrix}$

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