Math, asked by ashoksharma3876, 5 months ago

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Answers

Answered by tomaraditya65051
1

Step-by-step explanation:

arithmetic progression by using blur coons formul

Answered by joelpaulabraham
1

Step-by-step explanation:

Let the 1st term of the AP be a and its common difference be d.

Now, we are given,

a(pth) = q

a(qth) = p

Now, to find the nth term, we use the formula,

a(nth) = a + (n - 1)d

Thus,

a(pth) = a + (p - 1)d

∴ a + (p - 1)d = q ------ 1

Similarly,

a(qth) = a + (q - 1)d

∴ a + (q - 1)d = p ------ 2

Subtracting eq.1 and eq.2 we get,

a + (p - 1)d = q

- a + (q - 1)d = p

(-) (-) (-)

———————————

0 + (p - 1)d - (q - 1)d = q - p

———————————

Hence,

(p - 1)d - (q - 1)d = q - p

Taking d common,

d((p - 1) - (q - 1)) = q - p

Opening the brackets,

d(p - 1 - q + 1) = q - p

d(p - q) = q - p

d = (q - p)/(p - q)

Here,

p - q = (-1) × (q - p)

To check,

Using Distributive Property,

(-1) × q - (-1) × p

-q - (-p)

-q + p

p - q

Thus,

d = (q - p)/(-1) × (q - p)

d = 1/(-1) × 1

d = 1/(-1)

d = -1 ---- 3 [Because negatives can't be put in the numerators]

Now, Putting d = -1 in eq.1

a + (p - 1)(-1) = q

Using Distributive Property,

a + [(p × (-1)) - ((1) × (-1))] = q

a + (-p - (-1)) = q

a + (-p + 1) = q

a + 1 - p = q

a = p + q - 1 ------- 4

Hence,

a(nth) = a + (n - 1)d

From eq.3 and eq.4 we get,

a(nth) = (p + q - 1) + (n - 1)(-1)

Using Distributive Property,

a(nth) = p + q - 1 + [(n × (-1)) - (1 × (-1))]

a(nth) = p + q - 1 + (-n - (-1))

a(nth) = p + q - 1 + (1 - n)

a(nth) = p + q - 1 + 1 - n

a(nth) = p + q - n

Hence proved.

It might be a bit confusing and difficult to understand, so please do refer the above image as well.

Hope it helped and believing you understood it........All the best

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