Pro subscription

#735139
  • Resolved Sahaj
    Rank Math pro

    Hi,

    I had a subscription of Rankmath which was promised for $49 per year. But in December I received an invoice for $59 which I could not renew.
    Please find attached the screenshot of the email I was sent by Bhanu Ahluwalia.

    https://imgur.com/S8bo4vR

    Email received last year, from which I subscribed.

Viewing 4 replies - 1 through 4 (of 4 total)
  • Hello,

    I’m sorry to hear about the trouble you’ve faced with your Rank Math subscription. Unfortunately, I’m unable to view the screenshot you’ve attached. However, if you have any billing-related concerns or issues with your subscription, please feel free to reach out to our support team directly from here. Our team is available 24/7 to assist you with any billing-related queries.

    If you need further assistance, please don’t hesitate to ask.

    Best regards

    Sources:
    Rank Math PRO – Pricing Plans for Businesses & Individuals
    How To Upgrade Your Rank Math PRO Subscription » Rank Math
    How to Cancel Your Rank Math PRO Subscription » Rank Math

    Hello,

    Thank you for contacting Rank Math support.

    Since your subscription has expired, please purchase a new one from our site at the regular price. Once done, let us know and we will make a partial refund to match the price you were offered.

    Looking forward to helping you.

    Hi,

    I have subscribed to the normal Pro plan. Please change my plan to $49. Please make sure all future invoices are raised for $49 only. I did not renew last time for the same reason as, after committing to a lifetime price you are changing the price at the time of renewal.

    Hello,

    We just wanted to let you know that we have made the change and refunded $20 back to your payment source.

    The same shall be reflected in your payment source in 5-7 business days.

    The next renewal is going to be just $49 as well.

    We really look forward to hearing back from you. Thank you.

Viewing 4 replies - 1 through 4 (of 4 total)

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