Question

For what value of a, are (2a-1), 7 and 3a three consecutive terms of an A.P?​

Answer 1

Step-by-step explanation:

if the terms are in ap it means

2b=a+b

7=(2a-1)+3a

7+1=5a

a=8/5

so terms are

11/5,7,24/5

Answer 2

✨Given:✨

Three terms

• First term = (2a-1)
• Second Term = 7
• Third Term = 3a

✨To Find:✨

Value of a for which the above three terms are in an AP

✨Concept:✨

For three terms to be in AP , the sum of first and third term should be equal to twice the second term.

For three terms a , b and c to be in an AP:

2b = a + c

✨Answer:✨

From using the above concept, we get:

2 * 7 = 2a - 1 + 3a

5a - 1 = 14

5a = 14 + 1

5a = 15

a = $\dfrac{15}{5}$

a = 3

Therefore, the value of a is 3.

First Term = 2*3 - 1 = 6 - 1 = 5

Second Term = 7

Third Term = 3*3 = 9

✨Proof:✨

Substituting the value of a in the terms, we get:

First Term = (2*3 - 1) = 5

Second Term = 7

Third Term = 3*3 = 9

2b = a + c

2*7 = 5 + 9

14 = 14

LHS = RHS

Hence Proved.

✨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 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⭐

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

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

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