Question

if the sum of the first P terms of an AP is the same as the sum of its first Q terms (where p is not equal to Q) then show that the sum of its first (P + Q) terms is zero.?

Answer 1

Step-by-step explanation:

Next Gurukul

Learner's Section

WikiK-12 Wiki

AssessmentAssessments

ForumQ&A Forum

Educator's Section

MOOCMOOCs

Join Experts PanelJoin Experts Panel

Next WorldThe Next World

arrow_backAcademic Questions & Answers Forum

Sumit mahesh

Oct 11, 2013

If the sum of first p terms of an AP

If the sum of first p terms of an AP is equal to the sum of the first q terms then show that the sum of its first (p+q) term is 0 where p is not equal to q? Please give me answer as fast as you can...

question_answer Answers(2)

edit Answer

person

Agam Gupta

Member since Sep 21, 2013

Sum of first p terms of an AP is� Sum of first q terms of an AP is� Multiply both sides with (p + q)/2 Thus the sum of first (p + q) terms is zero

Answer 2

${\green {\boxed {\mathtt {✓verified\:answer}}}}$

$let \: \: a \: \: be \: the \: first \: term \: and \: d \: be \: the \: common \: difference \: of \: the \: given \: ap \: \\ then \\ s _{p} = s _{q} \implies \frac{p}{2} (2a + (p - 1)d) = \frac{q}{2} (2a + (q - 1)d \\ \implies(p - q)(2a) = (q - p)(q + p - 1) \\ \implies2a = (1 - p - q)d \: \: \: \: \: \: \: .....(1) \\ sum \: of \: the \: first \: (p + q) \: terms \: of \: the \: given \: ap \\ = \frac{(p + q)}{2} (2a + (p + q - 1)d) \\ = \frac{(p + q)}{2} .(1 - p - q)d + (p + q - 1)d \: \: \: \: \: \: \: \: (using \: 1) \\ = 0$

${\huge{\underline{\underline{\underline{\orange{\mathtt{❢◥ ▬▬▬▬▬▬ ⊙◆ ⊙▬▬▬▬▬▬ ◤❢}}}}}}}$