Question

Find the 32nd term of an A.P. whose first term and 20th term are respectively 2 and 59.​

Answer 1

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Answer 2

Given: The first and twentieth terms of an AP are 2 and 59.

To find: The thirty second term.

Answer:

The first term is 2.

⇒ a [the first term]  = 2.

The twentieth term is 59.

⇒ $\sf a_2_0\ =\ a\ +\ 19d\ =\ 59$

Subsituting the value of a into the equation above,

$\sf a_2_0\ =\ 2\ +\ 19\ \times\ d\\\\\\59\ =\ 2\ +\ 19d\\\\\\59\ -\ 2\ =\ 19d\\ \\\\57\ =\ 19d\\\\\\\dfrac{57}{19}\ =\ d\\\\\\3\ =\ d$

We now know that the common difference [d] is 3. Let's find the thirty second term.

$\sf a_3_2\ =\ a\ +\ 31d\\\\\\a_3_2\ =\ 2\ +\ 31\ \times\ 3\\\\\\a_3_2\ =\ 2\ +\ 93\\\\\\a_3_2\ =\ 95$

Therefore, the thirty second term of the AP is 95.