Question

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)​

Answer 1

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 ]

___________________

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) ]

________________________

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.)

Answer 2

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 :

$\dashrightarrow {\boxed{\tt{d \: = \: a_2 \: - \: a_1}}} \\ \\ \dashrightarrow \tt{d \: = \: b \: - \: a \: ----(1)}$

$\rule{150}{0.5}$

We also know that :

$\dashrightarrow {\boxed{\tt{a_n \: = \: a \: + \: (n \: - \: 1)d}}} \\ \\ \dashrightarrow \tt{c \: = \: a \: + \: (n \: - \: 1)(b \: - \: a)} \\ \\ \dashrightarrow \tt{c \: = \: a \: + \: bn \: - \: b \: - \: an \: + \: a} \\ \\ \dashrightarrow \tt{c \: = \: 2a \: + \: bn \: - \: an \: - \: b} \\ \\ \dashrightarrow \tt{c \: = \: n(b \: - \: a) \: - \: b \: + \: 2a} \\ \\ \dashrightarrow {\boxed{\boxed{\tt{n \: = \: \dfrac{c \: + \: b \: - \: 2a}{(b \: - \: a)}}}}}$

Now, use formula for sum of terms

$\dashrightarrow {\boxed{\tt{S_n \: = \: \dfrac{n}{2} (a \: \: + \: a_n)}}} \\ \\ \dashrightarrow \tt{S_n \: = \: \bigg( \dfrac{\dfrac{c \: + \: b \: - \: 2a}{b \: - \: a}}{2} \bigg) ( a \: + \: c )} \\ \\ \dashrightarrow {\boxed{\boxed{\tt{S_n \: = \: \dfrac{(c \: + \: b \: - \: 2a) (a \: + \: c)}{2(b \: - \: a)}}}}}$

HENCE PROVED