Question

Can someone please solve this and show the workings :
In an ap, the sum of the first 10 terms is 400 and the sum of the
Next 10 terms is 1000, find the common difference and the first term

Answer 1

Answer:

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

NOW,   20/2 [2a + (20-1)] = 400

 10[ 2a + 19d ]  = 400

     2a + 19d  =  40         ---------   (i)

similarly,   For 40 terms 

     40/2[ 2a+39d]  = 1600

       2a + 39d = 80           --------- (ii)

now on subtracting  (i) & (ii)

we get,    20d = 40          

                d =   2

 and therefore,    2a+ 19*2 = 40

                              a = 1

hence sum of 1st ten terms is

       10/2[2*1 + (10-1)2]

        5[ 2 + 18 ] = 20*5

          =  100

Answer 2

Answer:

Given, in an AP the sum of the first 10 terms is 400 and sum of the next 10 terms is 1000

Step-by-step explanation:

S10=400                                    ;         S20=400+1000=1400

Sn=n/2[2a+(n-1)d]                      ;         Sn=n/2[2a+(n-1)d]

400=10/2[2a+(10-1)d]                ;         1400=20/2[2a+(20-1)d]

400=5[2a+9d]                           ;          1400=10[2a+19d]

400/5=2a 9d                            ;         1400/10=2a+19d

80=2a+9d                                 ;         140=2a+19d

2a+9d=80-------(1)                     ;        2a+19d=140--------(2)

solving (1) and (2)

subtracting (1) and (2)

  2a+19d=140

  2a+9d=80

 0a+10d=60

    ⇒ 10d=60

    ⇒  d=60/10

    ⇒  d=6--------(3)

sub. d=6 in eq. (1)

2a+9d=80

2a+9(6)=80

2a+54=80

2a=80-54

2a=26

a=26/2

a=13