Question

for an ap t12 = 64 t20 = 112 find an ap​

Answer 1

Given:

$t_{12} = 64 \: \\ t_{20} = 112$

To Find:

The terms of AP.

Answer:

The General term of an AP is given by the formula :

$t_n = a + (n - 1)d$

The $12_{th}$ term is given as:

$t_{12} = a + (12 - 1)d \: \: \: \: \\ 64= a + 11d \: \: ......(1)$

Similarly,

The $20_{th}$ term is given as:

$t_{20} = a + (20 - 1)d \\ 112 = a + 19d \: \: ....(2)$

Subtracting equation (1) from (2), we get:

a + 19d - a - 11d = 48

8d = 48

$d = \dfrac{48}{8} \\ d = 6 \: \: \: \:$

Substituting the value of d in equation (1) , we get:

a + 11*6 = 64

a + 66 = 64

a = 64 - 66

a = -2

AP = -2 , 4 , 10 , 16 , ....

Other AP Formulas:

nth term of an AP formulas

$\\\\\sf 1) \: n_{th} \: term \: of \: any \: AP \: = a + (n - 1)d$

$\sf 2) \:n_{th} \: term \: from \: the \: end \: of \: an \: AP \: = a + (m - n)d$

$\sf 3) \:n_{th} \: term \: from \: the \: end \: of \: an \: AP = l - (n - 1)d$

$\sf 4) \: Difference \: of \: two \: terms = (m - n)d$

where m and n is the position of the term in AP

$\sf 5) \:Middle\: term\:of\: a\: finite\: AP\:$

$\sf (i) \:\: If \: n \: is \: odd = \frac{n + 1}{2}\:th\:term$

$\sf (ii) \:\: If \: n \: is \: even = \frac{n}{2} \:th \: term \: and \: ( \frac{n}{2} + 1)th \: term$

Sum Formulas

$\sf 1) \: Sum \: of \: first \: n \: terms \: of \: an \:AP = \frac{n}{2} [ \: 2a + (n - 1)d \: ]$

$\sf 2) \: Sum \: of \: first \: n \: natural \: numbers = \frac{n(n + 1)}{2}$

$\sf 3) \: Sum \: of \: AP \: having \: last \: term = \frac{n}{2} [ \: a + l \: ]$