Question

किसी समांतर श्रेणी (m+1) पद दोगुना है
(n+1 )में पद का तो सिद्ध करो कि (3m+1) वाँ पद 2 गुना होगा (m+n+1) वे पद का ?

For any parallel category (m + 1) and the term is double (n + 1), prove that the term (3m + 1) will be doubled (m + n + 1) of that post ?

Answer 1

!! Hey Mate !!

your answer is --

Given,

T(m+1) = 2 { T(n+1) }

=> a + (m+1-1)d = 2 { a+(n+1-1)d}

=> a + dm = 2(a+nd)

=> a + dm = 2a + 2nd

=> a = d(m-2n) .......(1)

Now,

T(3m+1) = a + 3md

= dm - 2nd + md { from 1 }

= -2nd + 4md

= -2d ( n-2m) .......(2)


now,

2{T (m+n+1)} = 2{a+(m+n+1-1)d}

= 2(a+dm+dn)

= 2(dm-2nd+dm+dn) { from 1 }

= 2( 2dm - dn )

= -2d ( n-2m) .......(3)


from (2) & (3)


T(3m+1) = 2{T(m+n+1)}


hence proved



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【Hope it helps you 】
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Answer 2

Given For any parallel category (m + 1) and the term is double (n + 1).

= > T(m + 1) = 2[T(n + 1)]

We know that sum of n terms of an AP = a + (n - 1) * d  ----- (1)

= > a + (m + 1 - 1) * d =  2[a + (n - 1 - 1) * d]

= > a + md = 2[a + nd]

= > a + md = 2a + 2nd

= > a + md - 2a = 2nd

= > -a = 2nd - md

= > a = md - 2nd --------- (1)


Now,

= > T(3m + 1) = a + (3m + 1 - 1) * d

                      = md - 2nd + 3md

                      = 4md - 2nd

                      = d(4m - 2n)  -------- (2)


Now,

We need to prove that (3m + 1) will be doubled (m + n - 1).

= > d(4m - 2n) = 2[T(m + n - 1)]

                        = 2[(a + (m + n - 1 - 1) * d]

                        = 2[(a + (m + n) * d]

                        = 2[a + md + nd]

                        = 2[md - 2nd + md + nd]

                        = 2[2md - nd]

                        = 4md - 2nd

                        = d(4m - 2n).



Therefore (3m + 1) will be doubled (m + n + 1).  ------ > Hence proved!.


Hope this helps!