JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 1)
Let Sn denote the sum of first n-terms of an arithmetic progression. If S10 = 530, S5 = 140, then S20 $$-$$ S6 is equal to:
1862
1842
1852
1872
Explanation
Let first term of A.P. be a and common difference is d.
$$\therefore$$ $${S_{10}} = {{10} \over 2}\{ 2a + 9d\} = 530$$
$$\therefore$$ $$2a + 9d = 106$$ ..... (i)
$${S_5} = {5 \over 2}\{ 2a + 4d\} = 140$$
$$a + 2d = 28$$ ...... (ii)
From equation (i) and (ii), a = 8, d = 10
$$\therefore$$ $${S_{20}} - {S_6} = {{20} \over 2}\{ 2 \times 8 + 19 \times 10\} - {6 \over 2}\{ 2 \times 8 + 5 \times 10\} $$
$$ = 2060 - 198$$
$$ = 1862$$
$$\therefore$$ $${S_{10}} = {{10} \over 2}\{ 2a + 9d\} = 530$$
$$\therefore$$ $$2a + 9d = 106$$ ..... (i)
$${S_5} = {5 \over 2}\{ 2a + 4d\} = 140$$
$$a + 2d = 28$$ ...... (ii)
From equation (i) and (ii), a = 8, d = 10
$$\therefore$$ $${S_{20}} - {S_6} = {{20} \over 2}\{ 2 \times 8 + 19 \times 10\} - {6 \over 2}\{ 2 \times 8 + 5 \times 10\} $$
$$ = 2060 - 198$$
$$ = 1862$$
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