please solve it fastly it is very urgent
Answers
Step-by-step explanation:
arithmetic progression by using blur coons formul
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