Question

in an AP the sum of its first 7term is 49 and also the sum its 14 tem is 196 find the sum of its 21 term​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

We will use Sum of AP formula then either you can use substitution or elimination to solve it

Step-by-step explanation:

$Sn = \frac{n}{2} (2a + (n-1)d)$

S₇ = $\frac{7}{2}(2a + 6d)$

49 = $\frac{7}{2} (2a + 6d)$

49*2 = 2(a + 3d)

49 = a + 3d

49 - 3d = a  -----------1

S₁₄ = $\frac{14}{2} ( 2a + 13d)$

196 = 7(2a + 13d)

196/7 = 2a + 13d

28 = 2a + 13d ----------2

Put a = 49 - 3d in (2)

28 = 2(49-3d) + 13d

28 = 98 - 6d + 13d

-70 = 7d

d = -10

Put d = -10 in Equation (1)

49 - 3(-10) = a

49 +30 = a

a = 79

Now S₂₁ = $\frac{21}{2} (2a + 20d)$

= $\frac{21}{2}(2*79 + 20(-10))\\\frac{21}{2}(158 - 200)$

$\frac{21}{2} * -42\\21*-21\\-441$