if the sum of first m terms of an AP is the same as the sum of its first n terms , show that the sum of its first (m+n) terms is zero.
Answers
Answer:
Let aa be the first term and dd be the common difference of the given A.P. Then, S_m = S_nS
m
=S
n
.
\Longrightarrow \dfrac{m}{2} \{2a + (m - 1)d \} = \dfrac{n}{2} \{2a + (n - 1) d\}⟹
2
m
{2a+(m−1)d}=
2
n
{2a+(n−1)d}
\Longrightarrow 2a(m - n) + \{m (m - 1) - n(n - 1)\}d = 0⟹2a(m−n)+{m(m−1)−n(n−1)}d=0
\Longrightarrow 2a(m - n) + \{(m^2 - n^2) - (m - n)\}d = 0⟹2a(m−n)+{(m
2
−n
2
)−(m−n)}d=0
\Longrightarrow (m - n) \{2a + (m + n - 1)d \} = 0⟹(m−n){2a+(m+n−1)d}=0
\Longrightarrow 2a + (m + n - 1)d = 0⟹2a+(m+n−1)d=0 [\because m - n \neq 0∵m−n
=0] ...(i)
Now,
S_{m + n} = \dfrac{m + n}{2} \{ 2a + (m + n - 1)d \}S
m+n
=
2
m+n
{2a+(m+n−1)d}
\Longrightarrow S_{m + n} = \dfrac{m + n}{2} \times 0 = 0⟹S
m+n
=
2
m+n
×0=0 [Using (i)]
Step-by-step explanation: