The first and the last term of an A .P. are 4 and 81 respectively. If the common difference is 7, how many terms are there in the A .P. and what is their sum?
Sn = 5n2 + 3n
⇒ Sn – 1 = 5(n – 1)2 + 3(n – 1)
∴ Sn – Sn – 1 = (5n2 + 3n) – [5(n – 1)2 + 3(n – 1)}
⇒ an = (5n2 + 3n) – {5 (n2 – 2n + 1) + 3n – 3}
= (5n2 + 3n) – {5n2 – 10n + 5 + 3n –3}
= 5n2 + 3n – 5n2 + 7n – 2
= 10n – 2
Hence, required nth term is 10n – 2
Sn = 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
⇒ Sn – 1 = 5(n – 1)2 + 3(n – 1)
∴ Sn – Sn – 1 = (5n2 + 3n) – [5(n – 1)2 + 3(n – 1)}
⇒ an = (5n2 + 3n) – {5 (n2 – 2n + 1) + 3n – 3}
= (5n2 + 3n) – {5n2 – 10n + 5 + 3n –3}
= 5n2 + 3n – 5n2 + 7n – 2
= 10n – 2
Hence, required nth term is 10n – 2.
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