Question

If 7,2,a,b,-13 are in A.P., then find a and b.​

Answer 1

Answer:

a= -3 and b= -8

Step-by-step explanation:

diff is 2-7= -5 by the use of diff a and b can be find.

Answer 2

Answer:

a = -3

b = -8

Given:

7 , 2 , a , b , -13 are in AP

To Find:

Value of a and b.

Solution:

We know that,

General Term of an AP = a+(n-1)d

where,

a = first term

n = number of terms in an AP or the position of a term in an AP

d = common difference of AP

In the given AP,

a = 7

d = 2-7 = -5

n = 5

$n_{3rd}$ = a + (3-1)d

a = 7 + 2*(-5)

a = 7-10

a = -3

$n_{4th}$ = a +(4-1)d

b = 7 + 3*(-5)

b = 7 - 15

b = -8

Therefore, the answer is:

a = -3

b = -8

✨Other AP Formulas:✨

⭐nth term of an AP

nth term of an APformulas⭐

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

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

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

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

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

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

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

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