Question

plzzz....answer my questionnnn​

Answer 1

Answer:

Step-by-step explanation:

first term = a

second term = b

last term = c

so, common difference = (b-a)

now, let last term be the Nth term of the AP

L = a + (n-1) d

=> c = a + (n-1) (b-a)

=> c = a + nb -an -b +a

=> c = 2a -an +nb -b

=> c = 2a -b +nb -an

=> n = (b+c-2a) / (b-a)

now, 

Sn = n/2 (first term + last term)

=> Sn= (b+c-2a)(a+b)/2(b-a)

Answer 2

welcome to the concept of Ap

given :

a = first term and b = T3 = a + 2d 

and c = Tn = a + ( n - 1)d.

Now T3 - T1 = b - a⇒ 2d = b - a.d = ( b -a) / 2.

Substitute d in nth term we get

c = a + ( n - 1) d

⇒ (c - a) = (n - 1) ( b -a) / 2.

⇒ 2(c - a) / ( b -a) = (n - 1)∴ n = ( b + 2c - 3a) / ( b - a).

Sum of n terms Sn= ( n / 2) [ a + l ]

⇒ Sn = ( ( b + 2c - 3a) / 2( b - a).)

[ a + c ]  

[ substitute n and l ]

⇒ Sn = ( b + 2c - 3a)( a + c ) / 2( b - a).

⇒ Sn = ( c + a ) / 2( b - a)[ b - a + 2( c -a) ]

Sn = [(c + a) / 2 + (c2- a2) / (b-a)].

I hope it help you ❤️

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