Question

if the first term of an AP is a and second term is b and last term is c then show that sum.of all terms is (a+c)(b+c-2a)/2(b-a)​

Answer 1

Answer:

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Step-by-step explanation:

ट्रांसलेट ई फॉलोइंग सेंटेंस इनटू ए मीडियम आफ इंस्ट्रक्शन

स्टॉक डिस्प्रेड ऑफर कोरोनावायरस

Answer 2

$\huge\underline\red{Answer:}$

$\small\green{It \: is \: given \: that,}$

$\small\green{First \: term = a}$

$\small\green{ ^{a}2 = b}$

$\small\green{l = c}$

$\small\green{Now,}$

$\small\green{ ^{a}n = a + (n - 1)d}$

$\small\green{ ^{a}2 = a + (2 - 1)d}$

$\small\green{ =>b = a + d}$

$\small\green{ => d = b - a...(1)}$

$\small\green{ l = a + (n - 1)d}$

$\small\green{ => c = a + (n - 1)(b - a)(from \: 1)}$

$\small\green{ => c - a = (n - 1)(b - a)}$

$\small\green{ => (n - 1) = \frac{c - a}{b - a}}$

$\small\green{ => n = \frac{c - a}{b - a} + 1}$

$\small\green{ => n = \frac{c - a + b - a}{b - a}}$

$\small\green{n = \frac{c + b - 2a}{b - a}...(2)}$

$\small\green{Also,}$

$\small\green{ ^{S}n = \frac{n}{2}(a + l)}$

$\small\green{ = \frac{ \frac{c + b - 2a}{b - a} }{2} (a + c)...(from \: 2)}$

$\small\green{ = \frac{(c + b - 2a)(a + c)}{2(b - a)}}$

$\small\green{Here, the \: sum \: of \: all \: terms \: is}$

$\small\green{ \frac{(a + c)(b + c - 2a)}{2(b - a)}}$

$\huge\underline\red{Explanation:}$

$\small\underline\green{Hope \: this \: answer \: will \: help \: you.}$

$\small\underline\green{Plz \: mark \: me \: as \: a \: Brainliest.}$