Question 9 Find the sum to n terms in the geometric progression 1, -a, a^2, -a^3... ( if a ≠ -1 )
Class X1 - Maths -Sequences and Series Page 192
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1, -a , a² , -a³,............ ( if a ≠ -1)
here,
first term of GP(a) = -a
common ratio ( r) = a²/(-a) = -a < 1
so , we have to use formula,
S_n = a( 1 - rⁿ)/(1 - r) [ because r < 1 ]
= 1{ 1 - (-a)ⁿ}/{1 - (-a)}
= {1 - (-a)ⁿ}/(a + 1)
hence, sum of n terms = {1 - (-a)ⁿ}/(a + 1)
here,
first term of GP(a) = -a
common ratio ( r) = a²/(-a) = -a < 1
so , we have to use formula,
S_n = a( 1 - rⁿ)/(1 - r) [ because r < 1 ]
= 1{ 1 - (-a)ⁿ}/{1 - (-a)}
= {1 - (-a)ⁿ}/(a + 1)
hence, sum of n terms = {1 - (-a)ⁿ}/(a + 1)
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