Live practice

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

After substitution and simplification:

By putting a = 9, d = 8, S = 636 in the formula S = [2 ( 1) ],2 + −n a n d we get a quadratic
equation 4n2 + 5n – 636 = 0. On solving, we get n = 53 , 124− . Out of these two roots only
one root 12 is admissible.
5. n = 16, d = 8
3 6. n = 38, S = 6973 7. Sum = 1661
8. S51 = 5610 9. n2 10. (i) S15 = 525 (ii) S15 = – 465
11. S1 = 3, S2 = 4; a2 = S2 – S1 = 1; S3 = 3, a3 = S3 – S2 = –1,
a10 = S10 – S9 = – 15; an = Sn – Sn – 1 = 5 – 2n.
12. 4920

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

After substitution and simplification:

960

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

After substitution and simplification:

625

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

After substitution and simplification:

₹ 27750

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

After substitution and simplification:

Values of the prizes (in ₹) are 160, 140, 120, 100, 80, 60, 40.

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

After substitution and simplification:

234

The length of a semicircle of radius \(r\) is \(\pi r\). The radii form an AP with \(a=0.5\), \(d=0.5\), and \(n=13\).

Therefore,

The row counts form an AP with \(a=20\), \(d=-1\), and sum 200.

The value \(n=25\) is invalid because it would make later row counts non-positive. Hence \(n=16\). The top row has

The one-way distances are \(5,8,11,\ldots,32\) m. Each collection requires an outward and return journey, so

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

After substitution and simplification:

32nd term