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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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