Show that the Sum of an AP whose first
tem is a the second term b and the
last term c is equal to Catc)(b +6-20)
20b-a)
Answers
Arithmetic Progression (AP) :-
• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.
• The nth term of an AP is given as ;
T(n) = a + (n-1)•d , where a is the first term and d is the common difference.
• The common difference of an AP is given as ;
d = T(n) - T(n-1)
• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.
Also, [(n+1)/2]th term will be its middle term.
• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.
Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.
• The sum up to nth terms of an AP is given as ;
S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.
• The nth term of an AP is also given as ;
T(n) = S(n) - S(n-1)
• Sum of nth term of AP with First & Last term : (n/2) [ first term + Last Term ]
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Solution :-
→ First Term = a
→ Second Term = b
→ Last Term = c .
As Told Above , we know that,
→ d = T(n) - T(n-1)
So,
→ d = T₂ - T₁
→ d = (b - a) .
Than,
→ Tₙ = a + (n - 1)d
→ Tₙ = a + (n - 1)(b-a)
→ Tₙ = a + bn - b - an + a
→ c = bn - an - b + 2a
→ c = n(b - a) - b
→ (c + b - 2a) = n(b - a)
→ n = [ (c + b - 2a) / ( b - a) ]
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Now, we know That, sum of n terms of AP is :-
→ (n/2) [ first term + Last Term ]
we Have :-
→ first Term = a
→ Last Term = c
→ n = [ (c + b - 2a) / ( b - a) ]
Putting Values we get :-
→ Sₙ = [ {(c + b - 2a) / ( b - a)} / 2 ] * [ a + c ]
→ Sₙ = {(c + b - 2a)(a + c)} / {2(b - a)} (Ans.)
GiveN :
- First term (a) = a
- Second term = b
- last Term (an) = c
To ShoW :
- (c + b - 2a)(a + c)/ 2(b - a)
SolutioN :
As we know that :
We also know that :
Now, use formula for sum of terms