Question

Let x1, X2, X3,... be terms of an Ap, if x1+x2+... +xn /x1 +x2+... +xm = n2/m2, (n not equal to m) , then x8 / x23 = ??
1: 1/3
2: 3
3: 1/9
4: 9​

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

Let x1, X2, X3,... be terms of an Ap, if x1+x2+... +xn /x1 +x2+... +xm = n²/m², (n not equal to m) , then x8 / x23 = ?? (Nice Question.)

Sᴏʟᴜᴛɪᴏɴ :-

→ (x1+x2+... +xn) /(x1 +x2+... +xm) = n²/m²

→ (sum of first n terms of AP) / (sum of first m terms of AP) = n²/m²

Let first term of AP is a and common difference is d.

Than,

using Sₙ = (n/2)[2a + (n - 1)d],

we get,

→ (n/2)[2a + (n-1)d] / (m/2)[2a + (m-1)d] = n²/m²

→ n*[2a + (n-1)d] / m[2a + (m-1)d] = n²/m²

→ [2a + (n-1)d] / [2a + (m-1)d] = n/m

→ 2am + (nd-d)m = 2an + (md-d)n

→ 2am + ndm - dm = 2an + mdn - dn = 0

→ 2am - 2an -dm + dn = 0

→ 2a(m - n) -d(m - n) = 0

→ (m - n)(2a - d) = 0

we get,

→ m = n

or,

→ 2a = d .

But we have given that, m ≠ n .

Therefore ,

→ 2a = d .

So,

→ T₈ / T₂₃

→ {a + (8 - 1)d} / {a + (23 - 1)d}

→ {a + 7d} / {a + 22d}

Putting d = 2a now,

→ {a + 7*2a} / {a + 22*2a}

→ {a + 14a} / {a + 44a}

→ {15a} / {45a}

→ (1/3) (Option B.) (Ans.)