Let Sn denote the sum of first n-terms of an arithmetic progression. If S10 = 530, S5 = 140, then S20 $-$ S6 is equal to:
Solution
Let first term of A.P. be a and common difference is d.<br><br>$\therefore$ ${S_{10}} = {{10} \over 2}\{ 2a + 9d\} = 530$<br><br>$\therefore$ $2a + 9d = 106$ ..... (i)<br><br>${S_5} = {5 \over 2}\{ 2a + 4d\} = 140$<br><br>$a + 2d = 28$ ...... (ii)<br><br>From equation (i) and (ii), a = 8, d = 10<br><br>$\therefore$ $${S_{20}} - {S_6} = {{20} \over 2}\{ 2 \times 8 + 19 \times 10\} - {6 \over 2}\{ 2 \times 8 + 5 \times 10\} $$<br><br>$= 2060 - 198$<br><br>$= 1862$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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