Short Answer TypeSn = 3n2 –4n
⇒ Sn – 1= 3 (n – 1)2 –4 (n – 1)
∴ Sn – Sn –1 = (3n2 – 4n) – {3(n – 1)2 – 4 (n –1)}
⇒ an = (3n2 – 4n) – {3(n2 – 2n+ 1) – 4n ± 4} = 3n2 – 4n – {3n2 – 6n + 3 – 4n + 4}
= 3n2 – 4n – {3n2 – 10n + 7}
= 3n2 – 4n – 3n2 + 10n – 7
= 6n – 7
Hence, required nth term is 6n – 7.
If Sn, the sum of first n terms of an A .P. is given by Sn = 5n2 + 3n, then find its nthterm.
Sn = 5n2 + 3n
Find the sum of the :
(i) First 25 terms of an A .P. whose nth term is given by an= 7 – 3n
Long Answer TypeIf in sin A .P. the sum of m terms is equal to n and the sum of n terms is equal to m, then prove that the sum of (m + n) terms is –(m + n).
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