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Question

 The sum of first 10 terms of an AP is -150 and the sum of its next 10 terms is -550. Find the AP.
 


Solution

Let a be the first term and d be the common difference of the AP.

Then,

S subscript 10 equals negative 150 space g i v e n not stretchy rightwards double arrow 10 over 2 left parenthesis 2 a plus 9 d right parenthesis equals negative 150 not stretchy rightwards double arrow 5 left parenthesis 2 a plus 9 d right parenthesis equals negative 150 not stretchy rightwards double arrow left parenthesis 2 a plus 9 d right parenthesis equals negative 30 not stretchy rightwards double arrow left parenthesis 2 a plus 9 d right parenthesis equals negative 30 rightwards arrow left parenthesis i right parenthesis

It is given that the sum of its next 10 terms is – 550

Now

S20 = Sum of first 20 terms = Sum of first 10 terms + Sum of the next 10 terms = – 150 + (- 550 ) = -700


not stretchy rightwards double arrow 20 over 2 left parenthesis 2 a plus 19 d right parenthesis equals � 700 not stretchy rightwards double arrow 10 left parenthesis 2 a plus 19 d right parenthesis equals � 700 not stretchy rightwards double arrow 2 a plus 19 d equals � 70 rightwards arrow left parenthesis ii right parenthesis

Subtracting (i) from (ii), we get
2 a plus 19 d � left parenthesis 2 a plus 9 d right parenthesis equals � 70 � left parenthesis negative 30 right parenthesis not stretchy rightwards double arrow 10 d equals negative 40 not stretchy rightwards double arrow d equals negative 4

Putting d = -4 in (i), we get
2 a plus 9 cross times left parenthesis negative 4 right parenthesis equals � 30 not stretchy rightwards double arrow 2 a equals � 30 plus 36 equals 6 a equals 3 H e n c e space r e q u i r e d space A P space i s space 3 comma negative 1 comma negative 5 comma negative 9...
 

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