Question
In an A.P of which $$a$$ is the first term, if the sum of the first $$p$$ terms is zero, then the sum of the next $$q$$ term is:
Found 5 tutors discussing this question
Discuss this question LIVE
13 mins ago
Text solutionVerified
SP (Sum of first $$P$$ tens) $$= \dfrac {P}{2} [2a + (P - 1)d]$$
$$d\rightarrow$$ common differences
$$SP = 0$$ (Given)
$$\dfrac {P}{2} [2a + (P - 1) d] = 0\Rightarrow 2a + (P - 1) d = 0$$
$$d = \dfrac {-2a}{(P - 1)} .... (i)$$
Sum of next $$q$$ tens $$=$$ Sum of first $$(p + q) tens -$$ sum of first $$P\ tens$$
$$S_{q'} = S_{p + q} - S_{q}$$
$$= \dfrac {(p + q)}{2} [2a + (p + q - 1)d] - \dfrac {p}{2} [2a + (p - 1)d]$$
$$= a \times q + \dfrac {d}{2} ((p + q)(p + q - 1) - p(p - 1)d)$$
$$= aq + \dfrac {d}{2} [p(p + q - 1) + q (p + q - 1) - p(p - 1)]$$
$$= aq + \dfrac {d}{2} [p(p - 1) + pq + q (p + q - 1) - p(p - 1)]$$
$$= aq + \dfrac {1}{2} \left (\dfrac {-2a}{p - 1}\right ) [p + p + q - 1] q$$
$$= aq \left (1 - \dfrac {1}{(p - 1)} (2p + q - 1)\right ]$$
$$= aq \left [\dfrac {p - 1 - 2p - q + 1}{p - 1}\right ] = aq \left [\dfrac {-p - q}{p - 1}\right ]$$
$$S_{q'} = \dfrac {aq (p + q)}{p -1}$$.
Was this solution helpful?
150
Share
Report
One destination to cover all your homework and assignment needs
Learn Practice Revision Succeed
Instant 1:1 help, 24x7
60, 000+ Expert tutors
Textbook solutions
Big idea maths, McGraw-Hill Education etc
Essay review
Get expert feedback on your essay
Schedule classes
High dosage tutoring from Dedicated 3 experts
Practice questions from similar books
Question 1
Complete the following activity to find the sum of natural number from to which are divisible by Sum of numbers from to , which are divisible by
Stuck on the question or explanation?
Connect with our maths tutors online and get step by step solution of this question.
231 students are taking LIVE classes
Question Text | In an A.P of which $$a$$ is the first term, if the sum of the first $$p$$ terms is zero, then the sum of the next $$q$$ term is: |
Answer Type | Text solution:1 |
Upvotes | 150 |