Question

# The 14th term of an AP is twice its $$8^{th}$$ term. If its 6th term is -8, then find the sum of its first 20 terms.

Solution

## Given,$$\\ { T }_{ 6 }=-8$$$$a+(6-1)d=-8$$$$a+5d=-8$$$$a=-8-5d$$$${ T }_{ 14 }=2{ T }_{ 8 }$$$$a+13d=2(a+7d)$$$$a+13d=2a+14d$$$$a=-d$$$$-8-5d=-d$$$$-8=4d$$$$d=-2$$$$a=2$$$$s_{ 20 }=\cfrac { 20 }{ 10 } (2+2+(20-1)(-2))$$$$=10(4+19(-2))$$$$=10(4-38)$$$$=-340$$Mathematics

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