Question

4.The seventh term of an AS is 40 and 13th term is 80.
(a)what is the tenth term?
(b)what is the sum of first 19 terms?
(c)what is the sum of first 19 terms of the A.S with seventh term 42 and thirteenth term 82?



*What is the Answer *​

Answer 1

Answer:

A7 = 40

a + 6d = 40 (eqn 1)

A13 = 80

a + 12d = 80 (eqn 2)

eqn 2 - eqn 1

a + 12d = 80

-a - 6d = -40

--------------------------

6d = 40

d = 40/6 = 20/3

a + 6 × 20/3 = 40

a + 40 = 40

a = 40 - 40

a = 0

a) A10 = a + 9d = 0 + 9 × 20/3 = 3 × 20 = 60

b) S19 = n/2 [ 2a + (n - 1)d]

= 19/2 [ 2×0 + (19-1) 20/3]

= 19/2 [ 18×20/3]

= 19/2 × 120

= 19 × 60

= 1150

c) The question you gave isn't clear, restate the question so I can edit it.

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Answer 2

Answer:

Step-by-step explanation:

let a & d be the first term & common difference respectively

$t_{7}=40\\\\a+(n-1)d=40\\\\a+(7-1)d=40\\\\a+6d=40\\\\a=40-6d$

$t_{13}=80\\\\a+(n-1)d=80\\\\a+(13-1)d=80\\\\a+12d=80\\\\40-6d+12d=80\\\\6d=40\\\\d=\frac{20}{3}$

$a = 40-6(\frac{20}{3})\\\\a=40-40\\\\a=0$

a)

$t_{10}=a+(n-1)d\\\\t_{10}=0+(10-1)(\frac{20}{3})\\\\t_{10}=9(\frac{20}{3})\\\\t_{10}=60$

b)

$S_{n}=\frac{n}{2}[2a+(n-1)d]\\\\S_{19}=\frac{19}{2}[2(0)+(19-1)(\frac{20}{3})]\\\\S_{19}=\frac{19}{2}[(18)(\frac{20}{3})]\\\\S_{19}=19[60]\\\\S_{19}=1140$

c)

$t_{7}=42\\\\a+(n-1)d=42\\\\a+(7-1)d=42\\\\a+6d=42\\\\a=42-6d$

$t_{13}=82\\\\a+(n-1)d=82\\\\a+(13-1)d=82\\\\a+12d=82\\\\42-6d+12d=82\\\\6d=40\\\\d=\frac{20}{3}$

$a = 42-6(\frac{20}{3})\\\\a=42-40\\\\a=2$

$S_{n}=\frac{n}{2}[2a+(n-1)d]\\\\S_{19}=\frac{19}{2}[2(2)+(19-1)(\frac{20}{3})]\\\\S_{19}=\frac{19}{2}[4+(18)(\frac{20}{3})]\\\\S_{19}=\frac{19}{2}[4+120]\\\\S_{19}=\frac{19}{2}(160)\\\\S_{19}=1570$