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Exercise 5.3 · Q2

Q.If the first term of a GP is 5 and common ratio is (−5)(-5), then which term is 3125?

(a) 6th
(b) 8th
(c) 5th
(d) 4th
Puducherry CbseNCERTSubjective· 1mImportance★★★★★est
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✓ Free question

Solving an=5(−5)n−1=3125a_n=5(-5)^{n-1}=3125 shows the 5th term equals 3125 — option (c).

[!FORMULA] an=a rn−1a_n=a\,r^{n-1}

aa = first term, rr = common ratio, nn = term number.

  1. Given a=5, r=−5a=5,\ r=-5. We need the nn for which an=3125a_n=3125: 5(−5)n−1=31255(-5)^{n-1}=3125.
  2. Divide both sides by 5: (−5)n−1=31255=625(-5)^{n-1}=\dfrac{3125}{5}=625.
  3. Since 625=54625=5^4 is positive, and (−5)n−1=(−1)n−15n−1(-5)^{n-1}=(-1)^{n-1}5^{n-1}, we need (−1)n−1=+1(-1)^{n-1}=+1 (so n−1n-1 is even) and 5n−1=625=54⇒n−1=4⇒n=55^{n-1}=625=5^4 \Rightarrow n-1=4 \Rightarrow n=5. (n−1=4n-1=4 is indeed even, consistent.)
  4. Check: a5=5(−5)4=5×625=3125a_5=5(-5)^4=5\times625=3125 ✓.
✓Final answer

(c) The 5th term equals 3125.

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